Đáp án:
a) $\left[\begin{array}{l}x = \dfrac12\arccos\dfrac15-\dfrac{1}{2}\arccos\dfrac35+ k\pi\\x =-\dfrac{1}{2}\arccos\dfrac35 -\arccos\dfrac15 + k\pi\end{array}\right.\quad (k\in\Bbb Z)$
b) $x = k\dfrac{\pi}{2}\qquad (k\in\Bbb Z)$
Giải thích các bước giải:
a) $3\cos2x -4\sin2x = 1$
$\to \dfrac35\cos2x -\dfrac45\sin2x =\dfrac15$
Ta có: $\left(\dfrac35\right)^2 +\left(\dfrac45\right)^2 = 1$
Đặt $\begin{cases}\cos\alpha =\dfrac35\\\sin\alpha =\dfrac45\end{cases}\longrightarrow \alpha =\arccos\dfrac35$
Phương trình trở thành:
$\cos2x.\cos\alpha -\sin2x.\sin\alpha =\dfrac15$
$\to \cos(2x +\alpha) = \dfrac15$
$\to \left[\begin{array}{l}2x +\alpha = \arccos\dfrac15 + k2\pi\\2x +\alpha =-\arccos\dfrac15 + k2\pi\end{array}\right.$
$\to \left[\begin{array}{l}x = \dfrac12\arccos\dfrac15-\dfrac{\alpha}{2} + k\pi\\x =-\dfrac{\alpha}{2} -\arccos\dfrac15 + k\pi\end{array}\right.$
$\to \left[\begin{array}{l}x = \dfrac12\arccos\dfrac15-\dfrac{1}{2}\arccos\dfrac35+ k\pi\\x =-\dfrac{1}{2}\arccos\dfrac35 -\arccos\dfrac15 + k\pi\end{array}\right.\quad (k\in\Bbb Z)$
b) $\cos(2x +1)=\cos(2x -1)$
$\to \left[\begin{array}{l}2x +1 = 2x -1 + k2\pi\\2x +1 = 1 - 2x + k2\pi\end{array}\right.$
$\to 4x = k2\pi$
$\to x = k\dfrac{\pi}{2}\qquad (k\in\Bbb Z)$
