Решить уравнение cos(-2x)=-√2/2, cosx=1/2 x∈ [latex] \left[ \frac{- \pi }{2};0 ] [/latex]

[latex]\cos(-2x)=-\frac{\sqrt{2}}{2} \\\ \cos2x=-\frac{\sqrt{2}}{2} \\\ 2x=\pm \frac{3\pi}{4}+2\pi n \\\ x=\pm\frac{3 \pi}{8}+\pi n[/latex] [latex]- \frac{\pi}{2} \leq \frac{3 \pi}{8}+\pi n \leq 0 \\\ - \frac{1}{2}\leq \frac{3}{8}+n\leq 0 \\ - \frac{1}{2}-\frac{3}{8} \leq n \leq -\frac{3}{8} \\\ -\frac{7 }{8}\leq n \leq -\frac{3}{8} \\\ n\neq \\\ - \frac{\pi}{2} \leq- \frac{3 \pi}{8}+\pi n \leq 0 \\\ - \frac{1}{2}\leq -\frac{3}{8}+n\leq 0 \\ - \frac{1}{2}+\frac{3}{8} \leq n \leq \frac{3}{8} \\\ - \frac{1}{8} \leq n \leq \frac{3}{8} \\\ n=0: \ x=- \frac{3 \pi }{8} + \pi \cdot0=- \frac{3 \pi }{8}[/latex] Ответ: -3π/8 [latex]-\frac{\pi}{2}\leq\frac{\pi}{3}+2\pi n\leq0 \\\ -\frac{1}{2}\leq\frac{1}{3}+2n\leq0 \\\ -\frac{1}{2}-\frac{1}{3}\leq2n\leq-\frac{1}{3} \\\ -\frac{5}{6}\leq2n\leq-\frac{1}{3} \\\ -\frac{5}{12}\leq n\leq-\frac{1}{6} \\\ n\neq \\\ -\frac{\pi}{2}\leq-\frac{\pi}{3}+2\pi n\leq0 \\\ -\frac{1}{2}\leq-\frac{1}{3}+2n\leq0 \\\ -\frac{1}{2}+\frac{1}{3}\leq2n\leq\frac{1}{3} \\\ -\frac{1}{6}\leq2n\leq\frac{1}{3} \\\ -\frac{1}{12}\leq n\leq\frac{1}{6} \\\ n=0: \ x=-\frac{\pi}{3} +2 \pi \cdot0=-\frac{\pi}{3}[/latex] Ответ: -π/3