a. sin2x+cos2x= -√2sinx b. cos4x+cos6x+cos8x=0 Giúp mk vs ạ

Đáp án:

a) $\left[\begin{array}{l}x = -\dfrac{\pi}{12}+ k\dfrac{2\pi}{3}\\x =\dfrac{3\pi}{4}  + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$

b) $\left[\begin{array}{l}x = \dfrac{\pi}{12}+ k\dfrac{\pi}{6}\\x =\dfrac{\pi}{3}  + k\pi\\x = -\dfrac{\pi}{3} + k\pi\end{array}\right.\quad (k\in\Bbb Z)$

Giải thích các bước giải:

a) $\sin2x +\cos2x = -\sqrt2\sin x$

$\to \dfrac{\sqrt2}{2}\sin2x +\dfrac{\sqrt2}{2}\cos2x = -\sin x$

$\to \sin\left(2x +\dfrac{\pi}{4}\right) = \sin(-x)$

$\to \left[\begin{array}{l}2x +\dfrac{\pi}{4} = – x + k2\pi\\2x +\dfrac{\pi}{4} = \pi + x + k2\pi\end{array}\right.$

$\to \left[\begin{array}{l}x = -\dfrac{\pi}{12}+ k\dfrac{2\pi}{3}\\x =\dfrac{3\pi}{4}  + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$

b) $\cos4x +\cos6x +\cos8x = 0$

$\to \cos6x + 2\cos6x\cos2x = 0$

$\to \cos6x(2\cos2x +1)=0$

$\to \left[\begin{array}{l}\cos6x = 0\\\cos2x = -\dfrac12\end{array}\right.$

$\to \left[\begin{array}{l}6x = \dfrac{\pi}{2}+ k\pi\\2x =\dfrac{2\pi}{3}  + k2\pi\\2x = -\dfrac{2\pi}{3} + k2\pi\end{array}\right.$

$\to \left[\begin{array}{l}x = \dfrac{\pi}{12}+ k\dfrac{\pi}{6}\\x =\dfrac{\pi}{3}  + k\pi\\x = -\dfrac{\pi}{3} + k\pi\end{array}\right.\quad (k\in\Bbb Z)$