Question

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Answer 1

$1)\; cos2x=-1\\\\2x=\pi +2\pin \; ,\; n\in Z\\\\x=\frac{\pi}{2}+\pi n\; ,\; n\in Z\\\\2)\; 2cos\frac{x}{3}=\sqrt3\\\\cos\frac{x}{3}=\frac{\sqrt3}{2}\\\\\frac{x}{3}=\pm arccos \frac{\sqrt3}{2}+2\pi n=\pm \frac{\pi}{6}+2\pi n\; ,\; n\in Z\\\\x=\pm \frac{\pi}{2}+6\pi n\; ,n\in Z\\\\3)\; cos(2x-\frac{\pi}{4})=0\\\\2x-\frac{\pi}{4}=\frac{\pi}{2}+\pi n\; ,\; n\in Z\\\\2x=\frac{\pi}{2}+\frac{\pi}{4}+\pi n=\frac{3\pi}{4}+\pi n,\; n\in Z\\\\x=\frac{3\pi}{8}+\frac{\pi n}{2}\; ,\; n\in Z$