9.5 Solving trigonometric equations  (Page 9/10)

${\csc }^{2}t=3$

${\cos }^{2}x=\frac{1}{4}$

$2\sin \theta =\mathrm{-1}$

$\tan x\sin x+\sin \left(-x\right)=0$

$9\sin \omega -2=4{\sin }^2\omega$

$1-2\tan (\omega )={\tan }^2(\omega )$

$\sec x\cos x+\cos x-\frac{1}{\sec x}$

${\sin }^{3}x+{\cos }^{2}x\sin x$

${\sin }^{2}x+{\sec }^{2}x-1=\frac{\left(1-{\cos }^{2}x\right)\left(1+{\cos }^{2}x\right)}{{\cos }^{2}x}$

${\tan }^{3}x{\csc }^{2}x{\cot }^{2}x\cos x\sin x=1$

$\tan \left(\frac{7\pi }{12}\right)$

$\cos \left(\frac{25\pi }{12}\right)$

$\sin (\mathrm{70°})\cos (\mathrm{25°})-\cos (\mathrm{70°})\sin (\mathrm{25°})$

$\cos (\mathrm{83°})\cos (\mathrm{23°})+\sin (\mathrm{83°})\sin (\mathrm{23°})$

$\cos \left(4x\right)-\cos \left(3x\right)\cos x={\sin }^{2}x-4{\cos }^{2}x{\sin }^{2}x$

$\cos (3x)-{\cos }^{3}x=-\cos x{\sin }^{2}x-\sin x\sin (2x)$

$\frac{\tan \left(\frac{1}{2}x\right)+\tan \left(\frac{1}{8}x\right)}{1-\tan \left(\frac{1}{8}x\right)\tan \left(\frac{1}{2}x\right)}$

$\cos \left({\sin }^{-1}\left(0\right)-{\cos }^{-1}\left(\frac{1}{2}\right)\right)$

$\tan \left({\sin }^{-1}\left(0\right)+{\sin }^{-1}\left(\frac{1}{2}\right)\right)$

Find $\sin \left(2\theta \right),\cos \left(2\theta \right),$ and $\tan \left(2\theta \right)$ given $\cos \theta =-\frac{1}{3}$ and $\theta$ is in the interval $\left[\frac{\pi }{2},\pi \right].$

Find $\sin \left(2\theta \right),\cos \left(2\theta \right),$ and $\tan \left(2\theta \right)$ given $\sec \theta =-\frac{5}{3}$ and $\theta$ is in the interval $\left[\frac{\pi }{2},\pi \right].$

$\sin \left(\frac{7\pi }{8}\right)$

$\sec \left(\frac{3\pi }{8}\right)$

$\sin (2\beta ),\cos (2\beta ),\tan (2\beta ),\sin (2\alpha ),\cos (2\alpha ),\text{and }\tan (2\alpha )$

$\sin \left(\frac{\beta }{2}\right),\cos \left(\frac{\beta }{2}\right),\tan \left(\frac{\beta }{2}\right),\sin \left(\frac{\alpha }{2}\right),\cos \left(\frac{\alpha }{2}\right),\text{and }\tan \left(\frac{\alpha }{2}\right)$

$\frac{2\cos \left(2x\right)}{\sin \left(2x\right)}=\cot x-\tan x$

$\cot x\cos (2x)=-\sin (2x)+\cot x$

${\cos }^{2}x{\sin }^4(2x)$

${\tan }^{2}x{\sin }^{3}x$

$\cos \left(\frac{\pi }{3}\right)\sin \left(\frac{\pi }{4}\right)$

$2\sin \left(\frac{2\pi }{3}\right)\sin \left(\frac{5\pi }{6}\right)$

$2\cos \left(\frac{\pi }{5}\right)\cos \left(\frac{\pi }{3}\right)$

$\sin \left(\frac{\pi }{12}\right)-\sin \left(\frac{7\pi }{12}\right)$

$\cos \left(\frac{5\pi }{12}\right)+\cos \left(\frac{7\pi }{12}\right)$

$\sin (9x)\cos (3x)$

$\cos (7x)\cos (12x)$

$\sin (11x)+\sin (2x)$

$\cos (6x)+\cos (5x)$

$\tan x+1=0$

$2\sin (2x)+\sqrt{2}=0$

$2{\sin }^{2}x-\sin x=0$

${\cos }^{2}x-\cos x-1=0$

$2{\sin }^{2}x+5\sin x+3=0$

$\cos x-5\sin \left(2x\right)=0$

$\frac{1}{{\sec }^{2}x}+2+{\sin }^{2}x+4{\cos }^{2}x=0$