1) Căn 3 sin2x + cos2x = 2cosx-1

Đáp án:

$x=\left\{{x=\dfrac{\pi}2+k\pi;k2\pi;\dfrac{2\pi}3+k2\pi}\right\}$

Giải thích các bước giải:

$\sqrt3\sin2x+\cos2x=2\cos x-1$

$\Leftrightarrow2\sqrt3\sin x\cos x+2\cos^2x-1=2\cos x-1$

$\Leftrightarrow\sqrt3\sin x\cos x+\cos^2x=\cos x$

$\Leftrightarrow\cos x(\sqrt3\sin x+\cos x-1)=0$

$\Leftrightarrow\left[\begin{array}{I}\cos x=0\\\sqrt3\sin x+\cos x=1\end{array}\right.$

$\Leftrightarrow\left[\begin{array}{I} x=\dfrac{\pi}2+k\pi\\\dfrac{\sqrt3}2\sin x+\dfrac12\cos x=\dfrac12\text{ (1)}\end{array}\right.$

(1) $\Leftrightarrow \cos\dfrac{\pi}6\sin x+\sin \dfrac{\pi}6\cos x=\dfrac12$

$\Leftrightarrow\sin\left({x+\dfrac{\pi}6}\right)=\dfrac12$

$\Leftrightarrow x+\dfrac{\pi}6=\dfrac{\pi}6+k2\pi$ hoặc $x+\dfrac{\pi}6=\pi-\dfrac{\pi}6+k2\pi$

$\Leftrightarrow x=k2\pi$ hoặc $x=\dfrac{2\pi}3+k2\pi$ $(k\in\mathbb Z)$

Vậy phương trình có nghiệm

$x=\left\{{x=\dfrac{\pi}2+k\pi;k2\pi;\dfrac{2\pi}3+k2\pi}\right\}$.