Question

помогите пожалуйста с системой тригонометрических уравнений

Answer 1

$x=\frac{4\pi}{3}+y\Rightarrow \sin^2\left ( \frac{4\pi}{3}+y \right )+\sin^2y=\frac{1}{2}\Leftrightarrow\left ( \cos y\sin \frac{4\pi}{3}+\cos \frac{4\pi}{3}\sin y \right )^2+\sin^2y=\frac{1}{2}$$\left (-\frac{\sqrt{3}}{2}\cos y-\frac{1}{2}\sin y \right )^2+\sin^2y=\frac{1}{2}\Leftrightarrow \left (\frac{\sqrt{3}}{2}\cos y+\frac{1}{2}\sin y \right )^2+\sin^2y=\frac{1}{2}$$\left ( \cos \frac{\pi}{6}\cos y+\sin \frac{\pi}{6}\sin y \right )^2+\sin^2y=\frac{1}{2}\Leftrightarrow \cos^2\left ( \frac{\pi}{6}-y \right ) +\sin^2y=\frac{1}{2}$$\cos\left ( 2y-\frac{\pi}{3} \right )-\cos 2y=1\Leftrightarrow 2\cos^2\left ( \frac{1}{2}\left ( 2y-\frac{\pi}{3} \right ) \right )-\cos 2y=0$$2\left ( \cos y\cos \frac{\pi}{6}+\sin y\sin \frac{\pi}{6} \right )^2-\cos 2y=0\Leftrightarrow 2\left ( \frac{1}{2}\sin y+\frac{\sqrt{3}}{2}\cos y \right )^2-c\cos 2y=0$$-\cos 2y+\frac{1}{2}\sin^2y+\sqrt{3}\cos y\sin y+\frac{3}{2}\cos^2y=0$$\frac{\sqrt{3}}{2}\sin 2y+\frac{1}{2}\left ( \cos 2y+1 \right )-\cos 2y+\frac{1}{2}=0\Leftrightarrow \frac{3}{2}\frac{\sin^2y}{\cos^2y}+\sqrt{3}\frac{\sin y}{\cos y}+\frac{1}{2}=0$$\frac{3}{2}\mathrm{tg}^2y+\sqrt{3}\mathrm{tg}y+\frac{1}{2}=0\Leftrightarrow \left ( \sqrt{\frac{3}{2}}\mathrm{tg}y+\frac{1}{\sqrt{2}} \right )^2=0$$\mathrm{tg}y=-\frac{1}{\sqrt{3}}\Rightarrow y=-\frac{\pi}{6}+\pi k,k\in \mathbb{Z}\Rightarrow x=\frac{7\pi}{6}+\pi k,k\in \mathbb{Z}$

Answer 2

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