Đáp án: $x=\dfrac{\pi }{6}+2\pi n,\:x=\dfrac{5\pi }{6}+2\pi n\:, n\in N$
Giải thích các bước giải:
Ta có:
$\cos(2x+\dfrac{\pi}{4})+\cos(2x-\dfrac{\pi}{4})+4\sin x=2+\sqrt{2}(1-\sin x)$
$\to 2\cos(\dfrac{2x+\dfrac{\pi}{4}+2x-\dfrac{\pi}{4}}{2})\cos(\dfrac{2x+\dfrac{\pi}{4}-(2x-\dfrac{\pi}{4})}{2})+4\sin x=2+\sqrt{2}(1-\sin x)$
$\to 2\cos(2x)\cos(\dfrac{\pi}{4})+4\sin x=2+\sqrt{2}(1-\sin x)$
$\to \sqrt{2}\cos(2x)+4\sin x=2+\sqrt{2}(1-\sin x)$
$\to \sqrt{2}(1-2\sin^2x)+4\sin x=2+\sqrt{2}(1-\sin x)$
Đặt $\sin x=t, -1\le t\le 1$
$\to \sqrt{2}(1-2t^2)+4t=2+\sqrt{2}(1-t)$
$\to -2\sqrt{2}t^2+\left(4+\sqrt{2}\right)t-2=0$
$\to t=\dfrac{1}{2},\:t=\sqrt{2}$
Mà $-1\le t\le 1\to t=\dfrac12$
$\to \sin x=\dfrac12$
$\to x=\dfrac{\pi }{6}+2\pi n,\:x=\dfrac{5\pi }{6}+2\pi n\:, n\in N$
