Question

Уравнение логарифмическое

Answer 1

$\displaystyle 2log_2(cos2x-sinx+2\sqrt{2})=3\\\\ODZ: cos2x-sinx+2\sqrt{2}>0\\\\ -1\leq cos2x\leq 1 ;\\\\ -1\leq sinx\leq 1; -1\leq -sinx\leq 1\\\\-2\leq cos2x-sinx\leq 2\\\\ 2\sqrt{2}>2$

значит

$\displaystyle -2+2\sqrt{2}\leq cos2x-sinx+2\sqrt{2}\leq 2+2\sqrt{2}\\\\0< cos2x-sinx+2\sqrt{2}$

решение

$\displaystyle log_2((1-2sin^2x)-sinx+2\sqrt{2})=\frac{3}{2}\\\\-2sin^2x-sinx+2\sqrt{2}+1=\sqrt{2}^3\\\\-2sin^2x-sinx+1=0\\\\sinx=t; |t|\leq 1\\\\-2t^2-t+1=0\\\\D=1+8=9\\\\t_{1.2}=\frac{1 \pm 3}{-4}\\\\t_1=-1; t_2=\frac{1}{2}$

$\displaystyle sinx=-1; x_1=\frac{3\pi }{2}+2\pi n; n \in Z\\\\sinx=\frac{1}{2}; x_2=\frac{\pi }{6}+2\pi n; n \in Z; x_3=\frac{5\pi }{6}+2\pi n; n \in Z$

выбор корней

$\displaystyle -\frac{5\pi }{2}; \frac{-11\pi }{6}; -\frac{7\pi }{6}$