Đáp án:
$86)$ vô nghiệm
$87) \,\, \left[\begin{array}{l}x = \dfrac{\pi}{6} + k\pi\\x =- \dfrac{\pi}{2} + k\pi\end{array}\right.\,\,\,(k\in \Bbb Z)$
Giải thích các bước giải:
$86) \, \, cos^2x - sinx = 2$
$\Leftrightarrow 1 - sin^2x - sinx - 2 = 0$
$\Leftrightarrow sin^2x + sinx + 1 = 0$
$\Leftrightarrow (sinx + \dfrac{1}{2})^2 +\dfrac{3}{4} = 0 \,\, (vô\,\, lí)$
Vậy phương trình đã cho vô nghiệm
$87)\,\, 4sin^2x + 3\sqrt{3}sin2x - 2cos^2x = 4$
$\Leftrightarrow 4.\left(\dfrac{1 - cos2x}{2}\right) + 3\sqrt{3}sin2x - 2.\left(\dfrac{1 + cos2x}{2}\right) = 4$
$\Leftrightarrow cos2x - \sqrt{3}sin2x = -1$
$\Leftrightarrow \dfrac{1}{2}cos2x -\dfrac{\sqrt{3}}{2}sin2x = -\dfrac{1}{2}$
$\Leftrightarrow sin\dfrac{\pi}{6}.cos2x - cos\dfrac{\pi}{6}.sin2x = sin\left(-\dfrac{\pi}{6}\right)$
$\Leftrightarrow sin\left(\dfrac{\pi}{6}-2x\right) = sin\left(-\dfrac{\pi}{6}\right)$
$\Leftrightarrow \left[\begin{array}{l}\dfrac{\pi}{6} - 2x = -\dfrac{\pi}{6} + k2\pi\\\dfrac{\pi}{6} - 2x = \dfrac{7\pi}{6} + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{6} + k\pi\\x =- \dfrac{\pi}{2} + k\pi\end{array}\right.\,\,\,(k\in \Bbb Z)$
