In Exercises solvebasicfirst - solvebasiclast, find all of the exact solutions of the equation and then list those solutions which are in the interval \([0, 2\pi )\).
\(\sin \left ( 5 \theta \right ) = 0\)

\(\theta = \frac {\pi k}{5}; \; \theta = 0, \frac {\pi }{5}, \frac {2\pi }{5}, \frac {3\pi }{5}, \frac {4\pi }{5}, \pi , \frac {6\pi }{5}, \frac {7\pi }{5}, \frac {8\pi }{5}, \frac {9\pi }{5}\)
\(\cos \left ( 3t \right ) = \frac {1}{2}\)

\(t = \frac {\pi }{9} + \frac {2\pi k}{3}\) or \(t = \frac {5\pi }{9} + \frac {2\pi k}{3}; \; t = \frac {\pi }{9}, \frac {5\pi }{9}, \frac {7\pi }{9}, \frac {11\pi }{9}, \frac {13\pi }{9}, \frac {17\pi }{9}\)
\(\sin \left ( -2x \right ) = \frac {\sqrt {3}}{2}\)

\(x = \frac {2\pi }{3} + \pi k\) or \(x = \frac {5\pi }{6} + \pi k; \; x = \frac {2\pi }{3}, \frac {5\pi }{6}, \frac {5\pi }{3}, \frac {11\pi }{6}\)
\(\tan \left ( 6 \theta \right ) = 1\)

\(\theta = \frac {\pi }{24} + \frac {\pi k}{6}; \; \theta = \frac {\pi }{24}, \frac {5\pi }{24}, \frac {3\pi }{8}, \frac {13\pi }{24}, \frac {17\pi }{24}, \frac {7\pi }{8}, \frac {25\pi }{24}, \frac {29\pi }{24}, \frac {11\pi }{8}, \frac {37\pi }{24}, \frac {41\pi }{24}, \frac {15\pi }{8}\)
\(\csc \left ( 4 t \right ) = -1\)

\(t = \frac {3\pi }{8} + \frac {\pi k}{2}; \; t = \frac {3\pi }{8}, \frac {7\pi }{8}, \frac {11\pi }{8}, \frac {15\pi }{8}\)
\(\sec \left ( 3x \right ) = \sqrt {2}\)

\(x = \frac {\pi }{12} + \frac {2\pi k}{3}\) or \(x = \frac {7\pi }{12} + \frac {2\pi k}{3}; \; x = \frac {\pi }{12}, \frac {7\pi }{12}, \frac {3\pi }{4}, \frac {5\pi }{4}, \frac {17\pi }{12}, \frac {23\pi }{12}\)
\(\cot \left ( 2 \theta \right ) = -\frac {\sqrt {3}}{3}\)

\(\theta = \frac {\pi }{3} + \frac {\pi k}{2}; \; \theta = \frac {\pi }{3}, \frac {5\pi }{6}, \frac {4\pi }{3}, \frac {11\pi }{6}\)
\(\cos \left ( 9t \right ) = 9\)

No solution
\(\sin \left ( \frac {x}{3} \right ) = \frac {\sqrt {2}}{2}\)

\(x = \frac {3\pi }{4} + 6\pi k\) or \(x = \frac {9\pi }{4} + 6\pi k; \; x = \frac {3\pi }{4}\)
\(\cos \left ( \theta + \frac {5\pi }{6} \right ) = 0\)

\(\theta = -\frac {\pi }{3} + \pi k; \; \theta = \frac {2\pi }{3}, \frac {5\pi }{3}\)
\(\sin \left ( 2t - \frac {\pi }{3} \right ) = -\frac {1}{2}\)

\(t = \frac {3\pi }{4} + \pi k\) or \(t = \frac {13\pi }{12} + \pi k; \; t = \frac {\pi }{12}, \frac {3\pi }{4}, \frac {13\pi }{12}, \frac {7\pi }{4}\)
\(2\cos \left ( x + \frac {7\pi }{4} \right ) = \sqrt {3}\)

\(x = -\frac {19\pi }{12} + 2\pi k\) or \(x = \frac {\pi }{12} + 2\pi k; \; x = \frac {\pi }{12}, \frac {5\pi }{12}\)
\(\csc ( \theta ) = 0\)

No solution
\(\tan \left ( 2t - \pi \right ) = 1\)

\(t = \frac {5\pi }{8} + \frac {\pi k}{2}; \; t = \frac {\pi }{8}, \frac {5\pi }{8}, \frac {9\pi }{8}, \frac {13\pi }{8}\)
\(\tan ^{2} \left ( x \right ) = 3\)

\(x = \frac {\pi }{3} + \pi k\) or \(x = \frac {2\pi }{3} + \pi k; \; x = \frac {\pi }{3}, \frac {2\pi }{3}, \frac {4\pi }{3}, \frac {5\pi }{3}\)
\(\sec ^{2} \left ( \theta \right ) = \frac {4}{3}\)

\(\theta = \frac {\pi }{6} + \pi k\) or \(\theta = \frac {5\pi }{6} + \pi k; \; \theta = \frac {\pi }{6}, \frac {5\pi }{6}, \frac {7\pi }{6}, \frac {11\pi }{6}\)
\(\cos ^{2} \left ( t \right ) = \frac {1}{2}\)

\(t = \frac {\pi }{4} + \frac {\pi k}{2}; \; t = \frac {\pi }{4}, \frac {3\pi }{4}, \frac {5\pi }{4}, \frac {7\pi }{4}\)
\(\sin ^{2} \left ( x \right ) = \frac {3}{4}\)

\(x = \frac {\pi }{3} + \pi k\) or \(x = \frac {2\pi }{3} + \pi k; \; x = \frac {\pi }{3}, \frac {2\pi }{3}, \frac {4\pi }{3}, \frac {5\pi }{3}\)
In Exercises solveidentfirst - solveidentlast, solve the equation, giving the exact solutions which lie in \([0, 2\pi )\)
\(\sin \left ( \theta \right ) = \cos \left ( \theta \right )\)

\(\theta = \frac {\pi }{4}, \frac {5\pi }{4}\)
\(\sin \left ( 2t \right ) = \sin \left ( t \right )\)

\(t = 0, \frac {\pi }{3}, \pi , \frac {5\pi }{3}\)
\(\sin \left ( 2x \right ) = \cos \left ( x \right )\)
\(x = \frac {\pi }{6}\) \(x = \frac {\pi }{3}\) \(x = \frac {\pi }{2}\) \(x = \frac {2\pi }{3}\) \(x = \frac {5\pi }{6}\) \(x = \pi \) \(x= \frac {3\pi }{2}\)
\(\cos \left ( 2\theta \right ) = \sin \left ( \theta \right )\)

\(\theta = \frac {\pi }{6}, \frac {5\pi }{6}, \frac {3\pi }{2}\)
\(\cos \left ( 2t \right ) = \cos \left ( t \right )\)
\(t=0\) \(t = \frac {\pi }{3}\) \(t = \frac {2\pi }{3}\) \(t = \pi \) \(t = \frac {4\pi }{3}\) \(t = \frac {5\pi }{3}\)
\(\cos (2x) = 2 - 5\cos (x)\)

\(x=\frac {\pi }{3}, \frac {5\pi }{3}\)
\(3\cos (2 \theta ) + \cos (\theta ) + 2 = 0\)

\(\theta = \frac {2\pi }{3}, \frac {4\pi }{3}, \arccos \left (\frac {1}{3}\right ), 2\pi -\arccos \left (\frac {1}{3}\right ) \)
\(\cos (2t) = 5\sin (t) - 2\)
\(t = \frac {\pi }{6}\) \(t = \frac {\pi }{3}\) \(t = \frac {\pi }{2}\) \(t = \frac {2\pi }{3}\) \(t = \frac {5\pi }{6}\) \(t = \pi \) \(t = \frac {3\pi }{2}\)
\(3\cos (2x) = \sin (x) + 2\)

\(x = \frac {7\pi }{6}, \frac {11\pi }{6}, \arcsin \left (\frac {1}{3}\right ), \pi - \arcsin \left (\frac {1}{3}\right ) \)
\(2\sec ^{2}(\theta ) = 3 - \tan (\theta )\)

\(\theta =\frac {3\pi }{4}, \frac {7\pi }{4}, \arctan \left (\frac {1}{2}\right ), \pi +\arctan \left (\frac {1}{2}\right ) \)
\(\tan ^{2}(t) = 1-\sec (t)\)
\(t=0\) \(t = \frac {\pi }{3}\) \(t = \frac {2\pi }{3}\) \(t = \pi \) \(t = \frac {4\pi }{3}\) \(t = \frac {5\pi }{3}\)
\(\cot ^{2}(x) = 3\csc (x) - 3\)
\(x = \frac {\pi }{6}\) \(x = \frac {\pi }{3}\) \(x = \frac {\pi }{2}\) \(x = \frac {2\pi }{3}\) \(x = \frac {5\pi }{6}\) \(x = \pi \)
\(\sec (\theta ) = 2\csc (\theta )\)

\(\theta =\arctan (2), \pi + \arctan (2)\)
\(\cos (t) \csc (t)\cot (t) = 6-\cot ^{2}(t)\)

\(t = \frac {\pi }{6}, \frac {7\pi }{6}, \frac {5\pi }{6}, \frac {11\pi }{6}\)
\(\sin (2x) = \tan (x)\)

\(x = 0, \pi , \frac {\pi }{4}, \frac {3\pi }{4}, \frac {5\pi }{4}, \frac {7\pi }{4}\)
\(\cot ^{4}(\theta ) = 4\csc ^{2}(\theta ) - 7\)

\(\theta = \frac {\pi }{6}, \frac {\pi }{4}, \frac {3\pi }{4}, \frac {5\pi }{6}, \frac {7\pi }{6}, \frac {5\pi }{4}, \frac {7\pi }{4}, \frac {11\pi }{6}\)
\(\cos (2t) + \csc ^{2}(t) = 0\)

\(t = \frac {\pi }{2}, \frac {3\pi }{2}\)
\(\tan ^{3} \left ( x \right ) = 3\tan \left ( x \right )\)

\(x = 0, \frac {\pi }{3}, \frac {2\pi }{3}, \pi , \frac {4\pi }{3}, \frac {5\pi }{3}\)
\(\tan ^{2} \left ( \theta \right ) = \frac {3}{2} \sec \left ( \theta \right )\)

\(\theta = \frac {\pi }{3}, \frac {5\pi }{3}\)
\(\cos ^{3} \left ( t \right ) = -\cos \left ( t \right )\)

\(t = \frac {\pi }{2}, \frac {3\pi }{2}\)
\(\tan (2x) - 2\cos (x) = 0\)

\(x = \frac {\pi }{6}, \frac {\pi }{2}, \frac {5\pi }{6}, \frac {3\pi }{2}\)
\(\csc ^{3}(\theta ) + \csc ^{2}(\theta ) = 4\csc (\theta ) + 4\)

\(\theta = \frac {\pi }{6}, \frac {5\pi }{6}, \frac {7\pi }{6}, \frac {3\pi }{2}, \frac {11\pi }{6}\)
\(2\tan (t) = 1 - \tan ^{2}(t)\)

\(t = \frac {\pi }{8}, \frac {5\pi }{8}, \frac {9\pi }{8}, \frac {13\pi }{8}\)
\(\tan \left ( x \right ) = \sec \left ( x \right )\)

No solution
In Exercises solvemoreidentfirst - solvemoreidentlast, solve the equation, giving the exact solutions which lie in \([0, 2\pi )\)
\(\sin (6\theta ) \cos (\theta ) = -\cos (6\theta ) \sin (\theta )\)

\(\theta = 0, \frac {\pi }{7}, \frac {2\pi }{7}, \frac {3\pi }{7}, \frac {4\pi }{7}, \frac {5\pi }{7}, \frac {6\pi }{7}, \pi , \frac {8\pi }{7}, \frac {9\pi }{7}, \frac {10\pi }{7}, \frac {11\pi }{7}, \frac {12\pi }{7}, \frac {13\pi }{7}\)
\(\sin (3t)\cos (t) = \cos (3t) \sin (t)\)

\(t=0, \frac {\pi }{2}, \pi , \frac {3\pi }{2}\)
\(\cos (2x)\cos (x) + \sin (2x)\sin (x) = 1\)

\(x = 0\)
\(\cos (5\theta )\cos (3\theta ) - \sin (5\theta )\sin (3\theta ) = \frac {\sqrt {3}}{2}\)

\(\theta = \frac {\pi }{48}, \frac {11\pi }{48}, \frac {13\pi }{48}, \frac {23\pi }{48}, \frac {25\pi }{48}, \frac {35\pi }{48}, \frac {37\pi }{48}, \frac {47\pi }{48}, \frac {49\pi }{48}, \frac {59\pi }{48}, \frac {61\pi }{48}, \frac {71\pi }{48}, \frac {73\pi }{48}, \frac {83\pi }{48}, \frac {85\pi }{48}, \frac {95\pi }{48}\)
\(\sin (t) + \cos (t) = 1\)
\(t = 0\) \(t = \frac {\pi }{6}\) \(t = \frac {\pi }{3}\) \(t = \frac {\pi }{2}\) \(t = \frac {2\pi }{3}\) \(t = \frac {5\pi }{6}\) \(t = \pi \)
\(\sin (x) + \sqrt {3} \cos (x) = 1\)

\(x = \frac {\pi }{2}, \frac {11\pi }{6}\)
\(\sqrt {2} \cos (\theta ) - \sqrt {2} \sin (\theta ) = 1\)

\(\theta = \frac {\pi }{12}, \frac {17\pi }{12}\)
\(\sqrt {3} \sin (2t) + \cos (2t) = 1\)
\(t=0\) \(t = \frac {\pi }{3}\) \(t = \frac {2\pi }{3}\) \(t = \pi \) \(t = \frac {4\pi }{3}\) \(t = \frac {5\pi }{3}\)
\(\cos (2x) - \sqrt {3} \sin (2x) = \sqrt {2}\)

\(x = \frac {17 \pi }{24}, \frac {41 \pi }{24}, \frac {23\pi }{24}, \frac {47\pi }{24}\)
\(3\sqrt {3}\sin (3\theta ) - 3\cos (3\theta ) = 3\sqrt {3}\)

\(\theta = \frac {\pi }{6}, \frac {5\pi }{18}, \frac {5\pi }{6}, \frac {17\pi }{18}, \frac {3\pi }{2}, \frac {29\pi }{18}\)
\(\cos (3t) = \cos (5t)\)

\(t = 0, \frac {\pi }{4}, \frac {\pi }{2}, \frac {3\pi }{4}, \pi , \frac {5\pi }{4}, \frac {3\pi }{2}, \frac {7\pi }{4}\)
\(\cos (4x) = \cos (2x)\)

\(x = 0, \frac {\pi }{3}, \frac {2\pi }{3}, \pi , \frac {4\pi }{3}, \frac {5\pi }{3}\)
\(\sin (5\theta ) = \sin (3\theta )\)

\(\theta = 0, \frac {\pi }{8}, \frac {3\pi }{8}, \frac {5\pi }{8}, \frac {7\pi }{8}, \pi , \frac {9\pi }{8}, \frac {11\pi }{8}, \frac {13\pi }{8}, \frac {15\pi }{8}\)
\(\cos (5t) = -\cos (2t)\)

\(t = \frac {\pi }{7}, \frac {\pi }{3}, \frac {3\pi }{7}, \frac {5\pi }{7}, \pi , \frac {9\pi }{7}, \frac {11\pi }{7}, \frac {5\pi }{3}, \frac {13\pi }{7}\)
\(\sin (6x) + \sin (x) = 0\)

\(x = 0, \frac {2\pi }{7}, \frac {4\pi }{7}, \frac {6\pi }{7}, \frac {8\pi }{7}, \frac {10\pi }{7}, \frac {12\pi }{7}, \frac {\pi }{5}, \frac {3\pi }{5}, \pi , \frac {7\pi }{5}, \frac {9\pi }{5}\)
\(\tan (x) = \cos (x)\)

\(x = \arcsin \left ( \frac {-1 + \sqrt {5}}{2} \right ) \approx 0.6662, \pi - \arcsin \left ( \frac {-1 + \sqrt {5}}{2} \right ) \approx 2.4754\)
In Exercises firstinveqn - lastinveqn, solve the equation.
\(\arccos (2x) = \pi \)

\(x = -\frac {1}{2}\)
\(\pi - 2\arcsin (t) = 2\pi \)

\(t=\answer {-1}\)

\(4\arctan (3x-1)-\pi =0\)

\(x = \frac {2}{3}\)
\(6 \, \text {arccot}(2t) - 5\pi = 0\)

\(t=\answer {-\frac {\sqrt {3}}{2}}\)

\(4 \,\text {arcsec}\left (\frac {x}{2}\right ) = \pi \)

\(x = 2\sqrt {2}\)
\(12 \,\text {arccsc}\left (\frac {t}{3}\right ) = 2\pi \)

\(t = \answer {6}\)

\(9 \arcsin ^{2}(x) - \pi ^2 = 0\)

\(x = \pm \frac {\sqrt {3}}{2}\)
\(9 \arccos ^{2}(t) - \pi ^2 = 0\)

\(t = \answer {\frac {1}{2}}\)

\(8 \, \text {arccot}^{2}(x)+3\pi ^2=10 \pi \, \text {arccot}(x)\)

\(x = -1,0\)
\(6 \arctan (t)^2= \pi \arctan (x)+\pi ^2\)

\(t = \answer {-\sqrt {3}}\)

In Exercises firstineqfirst - firstineqlast, solve the inequality. Express the exact answer in interval notation, restricting your attention to \(0 \leq x \leq 2\pi \).
\(\sin \left ( x \right ) \leq 0\)

\(\left [ \pi , 2\pi \right ]\)
\(\tan \left ( t \right ) \geq \sqrt {3}\)

\(\left [ \frac {\pi }{3}, \frac {\pi }{2} \right ) \cup \left [ \frac {4\pi }{3}, \frac {3\pi }{2} \right )\)
\(\sec ^{2} \left ( x \right ) \leq 4\)

\(\left [ 0, \frac {\pi }{3} \right ] \cup \left [ \frac {2\pi }{3}, \frac {4\pi }{3} \right ] \cup \left [ \frac {5\pi }{3}, 2\pi \right ]\)
\(\cos ^{2} \left ( t \right ) > \frac {1}{2}\)

\(\left [ 0, \frac {\pi }{4} \right ) \cup \left ( \frac {3\pi }{4}, \frac {5\pi }{4} \right ) \cup \left ( \frac {7\pi }{4}, 2\pi \right ]\)
\(\cos \left ( 2x \right ) \leq 0\)

\(\left [ \frac {\pi }{4}, \frac {3\pi }{4} \right ] \cup \left [ \frac {5\pi }{4}, \frac {7\pi }{4} \right ]\)
\(\sin \left ( t + \frac {\pi }{3} \right ) > \frac {1}{2}\)

\(\left [ 0, \frac {\pi }{2} \right ) \cup \left ( \frac {11\pi }{6}, 2\pi \right ]\)
\(\cot ^{2} \left ( x \right ) \geq \frac {1}{3}\)

\(\left ( 0, \frac {\pi }{3} \right ] \cup \left [ \frac {2\pi }{3}, \pi \right ) \cup \left ( \pi , \frac {4\pi }{3} \right ] \cup \left [ \frac {5\pi }{3}, 2\pi \right )\)
\(2\cos (t) \geq 1\)

\(\left [0, \frac {\pi }{3}\right ] \cup \left [\frac {5\pi }{3}, 2\pi \right ]\)
\(\sin (5x) \geq 5\)

No solution
\(\cos (3t) \leq 1\)

\([0, 2\pi ]\)
\(\sec (x) \leq \sqrt {2}\)

\(\left [0, \frac {\pi }{4} \right ] \cup \left (\frac {\pi }{2}, \frac {3\pi }{2}\right ) \cup \left [\frac {7\pi }{4}, 2\pi \right ]\)
\(\cot (t) \leq 4\)

\(\left [\text {arccot}(4), \pi \right ) \cup \left [ \pi + \text {arccot}(4), 2\pi \right )\)
In Exercises secondineqefirst - secondineqlast, solve the inequality. Express the exact answer in interval notation, restricting your attention to \(-\pi \leq x \leq \pi \).
\(\cos \left ( x \right ) > \frac {\sqrt {3}}{2}\)

\(\left ( -\frac {\pi }{6}, \frac {\pi }{6} \right )\)
\(\sin (t) > \frac {1}{3}\)

\(\left ( \arcsin \left (\frac {1}{3}\right ), \pi - \arcsin \left (\frac {1}{3}\right ) \right )\)
\(\sec \left ( x \right ) \leq 2\)

\(\left [ -\pi , -\frac {\pi }{2} \right ) \cup \left [ -\frac {\pi }{3}, \frac {\pi }{3} \right ] \cup \left ( \frac {\pi }{2}, \pi \right ]\)
\(\sin ^{2} \left ( t \right ) < \frac {3}{4}\)

\(\left ( -\frac {2\pi }{3}, -\frac {\pi }{3} \right ) \cup \left ( \frac {\pi }{3}, \frac {2\pi }{3} \right )\)
\(\cot \left ( x \right ) \geq -1\)

\(\left ( -\pi , -\frac {\pi }{4} \right ] \cup \left ( 0, \frac {3\pi }{4} \right ]\)
\(\cos (t) \geq \sin (t)\)

\(\left [ -\frac {3\pi }{4}, \frac {\pi }{4} \right ]\)
In Exercises thirdineqfirst - thirdineqlast, solve the inequality. Express the exact answer in interval notation, restricting your attention to \(-2\pi \leq x \leq 2\pi \).
\(\csc \left ( x \right ) > 1\)

\(\left ( -2\pi , -\frac {3\pi }{2} \right ) \cup \left ( -\frac {3\pi }{2}, -\pi \right ) \cup \left ( 0, \frac {\pi }{2} \right ) \cup \left ( \frac {\pi }{2}, \pi \right )\)
\(\cos (t) \leq \frac {5}{3}\)

\([-2\pi , 2\pi ]\)
\(\cot (x) \geq 5\)

\(\left (-2\pi , \text {arccot}(5) - 2\pi \right ] \cup \left (-\pi , \text {arccot}(5) - \pi \right ] \cup \left (0, \text {arccot}(5)\right ] \cup \left (\pi , \pi + \text {arccot}(5)\right ]\)
\(\tan ^{2} \left ( t \right ) \geq 1\)

\(\left [ -\frac {7\pi }{4}, -\frac {3\pi }{2} \right ) \cup \left ( -\frac {3\pi }{2}, -\frac {5\pi }{4} \right ] \cup \left [ -\frac {3\pi }{4}, -\frac {\pi }{2} \right ) \cup \left ( -\frac {\pi }{2}, -\frac {\pi }{4} \right ] \cup \left [ \frac {\pi }{4}, \frac {\pi }{2} \right ) \cup \left ( \frac {\pi }{2}, \frac {3\pi }{4} \right ] \cup \left [ \frac {5\pi }{4}, \frac {3\pi }{2} \right ) \cup \left ( \frac {3\pi }{2}, \frac {7\pi }{4} \right ]\)
\(\sin (2x) \geq \sin (x)\)

\(\left [ -2\pi , -\frac {5\pi }{3} \right ] \cup \left [ -\pi , -\frac {\pi }{3} \right ] \cup \left [ 0, \frac {\pi }{3} \right ] \cup \left [ \pi , \frac {5\pi }{3} \right ]\)
\(\cos (2t) \leq \sin (x)\)

\(\left [ -\frac {11\pi }{6}, -\frac {7\pi }{6} \right ] \cup \left [ \frac {\pi }{6}, \frac {5\pi }{6} \right ] \cup , \left \{ -\frac {\pi }{2}, \frac {3\pi }{2} \right \}\)
In Exercises invineqfirst - invineqlast, solve the given inequality.
\(\arcsin (2x) > 0\)

\(\left (0, \frac {1}{2}\right ]\)
\(3 \arccos (t) \leq \pi \)

\(\left [\frac {1}{2}, 1\right ]\)
\(6 \, \text {arccot}(7x) \geq \pi \)

\(\left (-\infty , \frac {\sqrt {3}}{7} \right ]\)
\(\pi > 2\arctan (t)\)

\((-\infty , \infty )\)
\(2\arcsin (x)^2 > \pi \arcsin (x)\)

\([-1,0)\)
\(12 \arccos (t)^2+2\pi ^2>11\pi \arccos (t)\)

\(\left [-1, -\frac {1}{2}\right ) \cup \left ( \frac {\sqrt {2}}{2}, 1\right ]\)
In Exercises domainfirst - domainlast, express the domain of the function using the extended interval notation. (See Example TrigDomainEx1 and Section extendedinterval for details.)
\(f(x) = \frac {1}{\cos (x) - 1}\)

\(\displaystyle \bigcup _{k=-\infty }^{\infty } \left ( 2k\pi , (2k+2)\pi \right )\)
\(f(t) = \frac {\cos (t)}{\sin (t) + 1}\)

\(\displaystyle \bigcup _{k=-\infty }^{\infty } \left ( \frac {(4k - 1)\pi }{2}, \frac {(4k + 3)\pi }{2} \right )\)
\(f(x) = \sqrt {\tan ^{2}(x) - 1}\)

\(\displaystyle \bigcup _{k=-\infty }^{\infty } \left \{ \left [ \frac {(4k + 1)\pi }{4}, \frac {(2k + 1)\pi }{2} \right ) \cup \left ( \frac {(2k + 1)\pi }{2}, \frac {(4k + 3)\pi }{4} \right ] \right \}\)
\(f(t) = \sqrt {2 - \sec (t)}\)

\(\displaystyle \bigcup _{k=-\infty }^{\infty } \left \{ \left [ \frac {(6k - 1)\pi }{3}, \frac {(6k + 1)\pi }{3} \right ] \cup \left ( \frac {(4k + 1)\pi }{2}, \frac {(4k + 3)\pi }{2} \right ) \right \}\)
\(f(x) = \csc (2x)\)

\(\displaystyle \bigcup _{k=-\infty }^{\infty } \left ( \frac {k\pi }{2}, \frac {(k+1)\pi }{2} \right )\)
\(f(t) = \frac {\sin (t)}{2 + \cos (t)}\)

\((-\infty , \infty )\)
\(f(x) = 3\csc (x) + 4\sec (x)\)

\(\displaystyle \bigcup _{k=-\infty }^{\infty } \left ( \frac {k\pi }{2}, \frac {(k+1)\pi }{2} \right )\)
\(f(t) = \ln \left ( |\cos (t)| \right )\)

\(\displaystyle \bigcup _{k=-\infty }^{\infty } \left ( \frac {(2k - 1)\pi }{2}, \frac {(2k+1)\pi }{2} \right )\)
\(f(x) = \arcsin (\tan (x))\)

\(\displaystyle \bigcup _{k=-\infty }^{\infty } \left [ \frac {(4k - 1)\pi }{4}, \frac {(4k+1)\pi }{4} \right ]\)
  1. With the help of your classmates, determine the number of solutions to \(\sin (x) = \frac {1}{2}\) in \([0,2\pi )\). Then find the number of solutions to \(\sin (2x) = \frac {1}{2}\), \(\sin (3x) = \frac {1}{2}\) and \(\sin (4x) = \frac {1}{2}\) in \([0,2\pi )\). What pattern emerges? Explain how this pattern would help you solve equations like \(\sin (11x) = \frac {1}{2}\).
  2. Repeat the above exercise focusing on \(\sin \left (\frac {x}{2}\right ) = \frac {1}{2}\), \(\sin \left (\frac {3x}{2}\right ) = \frac {1}{2}\) and \(\sin \left (\frac {5x}{2}\right ) = \frac {1}{2}\). What pattern emerges here?
  3. Replace sine with tangent and \(\frac {1}{2}\) with \(1\) and repeat the whole exploration.
Suppose an object weighing \(10\) pounds is suspended from the ceiling by a spring which stretches \(2\) feet to its equilibrium position when the object is attached.

  1. Find the spring constant \(k\) in \(\frac {\text {lbs.}}{\text {ft.}}\) and the mass of the object in slugs.

    \(k = 5 \, \frac {\text {lbs.}}{\text {ft.}}\) and \(m = \frac {5}{16} \, \text {slugs}\)

  2. Find the equation of motion of the object if it is released from \(1\) foot below the equilibrium position from rest. When is the first time the object passes through the equilibrium position? In which direction is it heading?

    \(x(t) = \sin \left (4t + \frac {\pi }{2}\right )\). The object first passes through the equilibrium point when \(t = \frac {\pi }{8} \approx 0.39\) seconds after the motion starts. At this time, the object is heading upwards.

  3. Find the equation of motion of the object if it is released from \(6\) inches above the equilibrium position with a downward velocity of \(2\) feet per second. Find when the object passes through the equilibrium position heading downwards for the third time.

    \(x(t) = \frac {\sqrt {2}}{2} \sin \left (4t + \frac {7\pi }{4}\right )\). The object passes through the equilibrium point heading downwards for the third time when \(t = \frac {17\pi }{16} \approx 3.34\) seconds.
In Example sinecurvesketckex01, \(f(x) = 2\sin (x) - \sin (2x)\) restricted to \(0 \leq x \leq 2\pi \). If \(f''(x) = 4 \sin (2x) - 2\sin (x)\), find the inflection points of the graph of \(y = f(x)\).

The inflection points are: \(\left ( \arccos \left (\frac {1}{4}\right ), \frac {3 \sqrt {15}}{8} \right )\), \(( \pi , 0)\), and \(\left ( 2\pi - \arccos \left (\frac {1}{4}\right ), - \frac {3 \sqrt {15}}{8} \right )\)
In Example IPexamplecosine, \(f(x) =x-2\cos (x)\) restricted to \(0 \leq x \leq 2\pi \). If \(f'(x) = 2\sin (x)+1\), list the open intervals over which \(f\) is increasing and decreasing. Find the local extrema.

\(f\) is increasing on \(\left (0, \frac {7\pi }{6} \right )\) and again on \(\left (\frac {11\pi }{6}, 2\pi \right )\); \(f\) is decreasing on \(\left (\frac {7 \pi }{6}, \frac {11\pi }{6} \right )\); local (absolute) max: \(\left (\frac {7\pi }{6}, \frac {7\pi }{6} + \sqrt {3}\right )\); local min: \(\left (\frac {11\pi }{6}, \frac {11\pi }{6} - \sqrt {3}\right )\)
Let \(f(x) = e^{-x} \, \sin (x)\) for \(x > 0\).

  1. Use the Squeeze Theorem, Theorem squeezeth, to find \(\ds {\lim _{x \rightarrow \infty } f(x)}\). Interpret your answer graphically.

    Since \(-1 \leq \sin (x) \leq 1\), \(-e^{-x} \leq e^{-x} \sin (x) \leq e^{-x}\) …

    \(\ds {\lim _{x \rightarrow \infty } f(x) = 0}\). We have a horizontal asymptote \(y = 0\).

  2. Use the fact that \(f'(x) = e^{-x} \cos (x) - e^{-x} \sin (x)\) to help you find the intervals over which \(f\) is increasing and decreasing.

    • \(f\) is increasing on \(\left (0, \frac {\pi }{4} \right )\), \(\left (\frac {5\pi }{4}, \frac {9 \pi }{4} \right )\), \(\left (\frac {13\pi }{4}, \frac {17 \pi }{4} \right )\), …:

      In other words: on \(\left (0, \frac {\pi }{4} \right )\) along with \(\left ( \frac {(8k+5) \pi }{4}, \frac {(8k+9)\pi }{4}\right )\) for \(k = 0, 1, 2, 3, \ldots \)

    • \(f\) is decreasing on\(\left (\frac {\pi }{4}, \frac {5 \pi }{4} \right )\), \(\left (\frac {9\pi }{4}, \frac {13 \pi }{4} \right )\), \(\left (\frac {17\pi }{4}, \frac {21\pi }{4} \right )\), …:

      In other words: on \(\left ( \frac {(8k+1) \pi }{4}, \frac {(8k+5)\pi }{4}\right )\) for \(k = 0, 1, 2, 3 \ldots \)

  3. Use the fact that \(f''(x) = -2 e^{-x} \cos (x)\) to help you find the intervals over which the graph of \(f\) is concave up and concave down.

    • the graph of \(f\) is concave up on \(\left (\frac {\pi }{2}, \frac {3\pi }{2} \right )\), \(\left (\frac {5\pi }{2}, \frac {7 \pi }{2} \right )\), \(\left (\frac {9\pi }{2}, \frac {11 \pi }{2} \right )\), …:

      In other words: on \(\left ( \frac {(4k+1) \pi }{2}, \frac {(4k+3)\pi }{2}\right )\) for \(k = 0,1, 2, 3, \ldots \)

    • the graph of \(f\) is concave down on \(\left (0, \frac {\pi }{2} \right )\), \(\left (\frac {3\pi }{2}, \frac {5 \pi }{2} \right )\), \(\left (\frac {7 \pi }{2}, \frac {9\pi }{2} \right )\), …:

      In other words: on \(\left (0, \frac {\pi }{2} \right )\) along with \(\left ( \frac {(4k+3) \pi }{2}, \frac {(4k+5)\pi }{2}\right )\) for \(k = 0, 1, 2, 3 \ldots \)

Recall \(x(t) = 10e^{-t/5} \sin \left (t + \frac {\pi }{3}\right )\) from Example underdampedresonance number underdampedproblem models underdamped motion. Use the Squeeze Theorem, Theorem squeezeth, to prove \(\ds {\lim _{t \rightarrow \infty } x(t) = 0}\).

HINT: Since \(-1 \leq \sin \left (t + \frac {\pi }{3}\right ) \leq 1\), \(-10e^{-t/5} \leq 10e^{-t/5} \sin \left (t + \frac {\pi }{3}\right ) \leq 10e^{-t/5}\) …