Question

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Cos 2x=1-sin2x
Tg^2 3x- 2 sin^2 3x=0
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Answer 1

$\displaystyle \cos 2x=1-\sin 2x\\ \cos^2x-\sin^2x-\cos^2x-\sin^2x+2\cos x\sin x=0\\ -2\sin^2x+2\cos x\sin x=0\\ -2\sin x(\sin x-\cos x)=0$

$-2\sin x=0\\ \sin x=0\\ x=\pi n,\;n\in Z$  или $\sin x - \cos x = 0 \;\;\;\;\; \mid:\sqrt{2} \\ \cos \frac{\pi}{4}\sin x-\sin \frac{\pi}{4}\cos x=0\\ \sin (x-\frac{\pi}{4})=0\\ x-\frac{\pi}{4}=\pi k, \; k\in Z\\ x=\frac{\pi}{4} +\pi k, \; k\in Z$

Ответ: $\pi n, n\in Z;\;\; \frac{\pi}{4}+\pi k, k\in Z$

$\displaystyle \tt tg^23x-2\sin^23x=0\\\\ \frac{\sin^2 3x}{\cos^2 3x}-2\sin^2 3x=0 \;\;\; |\cdot \cos^23x\\ \\ \sin^23x-2\sin^23x\cos^23x=0\\\\ 2\sin^23x(\cos^23x-\frac{1}{2})=0$

$\displaystyle \sin^23x=0 \\ \sin 3x=0 \\ 3x=\pi n, \; n\in Z \;\; |:3\\ x=\frac{\pi n}{3}, \; n\in Z$

или

$\displaystyle (\cos^23x-\frac{1}{2})=0\\ \left [{ {{\cos 3x=\frac{\sqrt{2}}{2}} \atop {\cos 3x=-\frac{\sqrt{2}}{2}}} \right. \left [{ {{3x=\frac{\pi}{4}+\pi m, \; m\in Z} \atop {3x=-\frac{\pi}{4}+\pi m, \; m\in Z}} \right. \left [{ {{x=\frac{\pi}{12}+\frac{\pi m}{3},\; m\in Z} \atop {x=-\frac{\pi}{12}+\frac{\pi m}{3}, \; m\in Z}} \right.$

Ответ: $\frac{\pi n}{3}, \; n\in Z; \;\; \frac{\pi }{12}+\frac{\pi m}{3}; \;\;-\frac{\pi}{12}+\frac{\pi m}{3}, \;m\in Z$