1) Cos^4x-sin^4x+5cosx+3=0 2) 3cos^2x+cos^2x×sinx=8(1+sinx)

Đáp án:

1) $x = \pm \dfrac{2\pi}{3}+k2\pi\quad (k \in \Bbb Z)$

2) $x = -\dfrac{\pi}{2} + k2\pi\quad (k\in \Bbb Z)$

Giải thích các bước giải:

1) $\cos^4x – \sin^4x + 5 \cos x + 3 = 0$

$\Leftrightarrow (\cos^2x – \sin^2x)(\cos^2x + \sin^2x) + 5 \cos x + 3 = 0$

$\Leftrightarrow 1.(2\cos^2x – 1)+ 5 \cos x + 3 = 0$

$\Leftrightarrow 2\cos^2x + 5\cos x + 2 = 0$

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

$\Leftrightarrow x = \pm \dfrac{2\pi}{3}+k2\pi\quad (k \in \Bbb Z)$

2) $3\cos^2x + \cos^2x.\sin x = 8(1+\sin x)$

$\Leftrightarrow 3(1 – \sin^2x) + (1-\sin^2x)\sin x = 8 + 8\sin x$

$\Leftrightarrow \sin^3x + 3\sin^2x + 7\sin x + 5 = 0$

$\Leftrightarrow \sin x = -1$

$\Leftrightarrow x = -\dfrac{\pi}{2} + k2\pi\quad (k\in \Bbb Z)$