Đáp án:

Giải thích các bước giải:

$b)\quad \cos x + \sqrt3\sin x = 2\cos\left(\dfrac{\pi}{3} - x\right)$

$\Leftrightarrow \dfrac12\cos x + \dfrac{\sqrt3}{2}\sin x = \cos\left(x - \dfrac{\pi}{3}\right)$

$\Leftrightarrow \cos\left(x - \dfrac{\pi}{3}\right)= \cos\left(x - \dfrac{\pi}{3}\right)$

Phương trình nghiệm đúng với mọi $x$

$c)\quad 2\cos\left(x + \dfrac{\pi}{6}\right) + 3\cos\left(x -\dfrac{\pi}{3}\right)= \dfrac{5\sqrt2}{2}$

$\Leftrightarrow 2\cos\left(x + \dfrac{\pi}{6}\right)  +3\cos\left(\dfrac{\pi}{3} - x\right)= \dfrac{5\sqrt2}{2}$

$\Leftrightarrow 2\cos\left(x + \dfrac{\pi}{6}\right)+  3\cos\left(\dfrac{\pi}{2} - \dfrac{\pi}{6} - x\right)= \dfrac{5\sqrt2}{2}$

$\Leftrightarrow 2\cos\left(x + \dfrac{\pi}{6}\right) + 3\sin\left(x +\dfrac{\pi}{3}\right)= \dfrac{5\sqrt2}{2}$

$\Leftrightarrow \dfrac{2}{\sqrt{13}}\cos\left(x + \dfrac{\pi}{6}\right) + \dfrac{3}{\sqrt{13}}\sin\left(x +\dfrac{\pi}{3}\right)= \dfrac{5\sqrt{26}}{26}$

Đặt $\begin{cases}\cos\alpha = \dfrac{2}{\sqrt{13}}\\\sin\alpha = \dfrac{3}{\sqrt{13}}\end{cases}\Rightarrow \alpha = \arccos\left(\dfrac{2}{\sqrt{13}}\right)$

Phương trình trở thành:

$\quad \cos\left(x + \dfrac{\pi}{6}\right)\cos\alpha + \sin\left(x + \dfrac{\pi}{6}\right)\sin\alpha = \dfrac{5\sqrt{26}}{26}$

$\Leftrightarrow \cos\left(x + \dfrac{\pi}{6} - \alpha\right)= \dfrac{5\sqrt{26}}{26}$

$\Leftrightarrow \left[\begin{array}{l}x + \dfrac{\pi}{6} - \alpha = \arccos\left(\dfrac{5\sqrt{26}}{26}\right) + k2\pi\\x + \dfrac{\pi}{6} - \alpha = - \arccos\left(\dfrac{5\sqrt{26}}{26}\right) + k2\pi\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l}x = - \dfrac{\pi}{6} +\alpha + \arccos\left(\dfrac{5\sqrt{26}}{26}\right) + k2\pi\\x = -\dfrac{\pi}{6} + \alpha - \arccos\left(\dfrac{5\sqrt{26}}{26}\right) + k2\pi\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l}x = - \dfrac{\pi}{6} +\arccos\left(\dfrac{2}{\sqrt{13}}\right)+ \arccos\left(\dfrac{5\sqrt{26}}{26}\right) + k2\pi\\x = -\dfrac{\pi}{6} + \arccos\left(\dfrac{2}{\sqrt{13}}\right) - \arccos\left(\dfrac{5\sqrt{26}}{26}\right) + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$

$d)\quad 2\sin^2x + \sqrt3\sin2x = 3$

$\Leftrightarrow 1 - \cos2x + \sqrt3\sin2x = 3$

$\Leftrightarrow \sqrt3\sin2x - \cos2x = 2$

$\Leftrightarrow \dfrac{\sqrt3}{2}\sin2x - \dfrac12\cos2x = 1$

$\Leftrightarrow \sin\left(2x - \dfrac{\pi}{6}\right)= 1$

$\Leftrightarrow 2x - \dfrac{\pi}{6}= \dfrac{\pi}{2} + k2\pi$

$\Leftrightarrow x = \dfrac{\pi}{3} + k\pi\quad (k\in\Bbb Z)$