Giải phương trình : 5sinx -2 = 3(1-sinx) tan²x

Đáp án:

$\left[\begin{array}{l}x = \dfrac{\pi}{6}+ k\pi\\x = \dfrac{5\pi}{6} + k2\pi\end{array}\right.(k\in\Bbb Z)$

Giải thích các bước giải:

$5\sin x - 2 = 3(1 - \sin x)\tan^2x \qquad (*)$

$ĐKXĐ:\, x \ne \dfrac{\pi}{2} + n\pi$

$(*)\Leftrightarrow 5\sin x - 2 = 3(1 - \sin x)\dfrac{\sin^2x}{\cos^2x}$

$\Leftrightarrow 5\sin x - 2 = 3(1 - \sin x)\dfrac{\sin^2x}{1 -\sin^2x}$

$\Leftrightarrow 5\sin x - 2 = 3(1 - \sin x)\dfrac{\sin^2x}{(1 -\sin x)(1 + \sin x)}$

$\Leftrightarrow 5\sin x - 2 = 3\dfrac{\sin^2x}{1 +\sin x}$

$\Leftrightarrow (5\sin x - 2)(1 + \sin x) = 3\sin^2x$

$\Leftrightarrow 2\sin^2x + 3\sin x - 2 = 0$

$\Leftrightarrow \left[\begin{array}{l}\sin x = \dfrac{1}{2}\\\sin x = -2\quad (loại)\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{6}+ k\pi\\x = \dfrac{5\pi}{6} + k2\pi\end{array}\right.(k\in\Bbb Z)$