Đáp án:
Giải thích các bước giải:
$a)\quad 2\sin^22x + \sqrt3\sin4x = - 3$
$\Leftrightarrow 1 - \cos4x + \sqrt3\sin4x = - 3$
$\Leftrightarrow \sqrt3\sin4x - \cos4x = - 4$
$\Leftrightarrow \sin\left(4x - \dfrac{\pi}{6}\right)= - 2$ (vô nghiệm )
Vậy phương trình đã cho vô nghiệm
$b)\quad 2\sin\left(x +\dfrac{\pi}{4}\right) + \sin\left(x - \dfrac{\pi}{4}\right)= \dfrac{3\sqrt2}{2}$
$\Leftrightarrow 2\sin\left(x +\dfrac{\pi}{4}\right) + \sin\left(\dfrac{\pi}{4}- x\right)= \dfrac{3\sqrt2}{2}$
$\Leftrightarrow 2\sin\left(x +\dfrac{\pi}{4}\right) + \sin\left(\dfrac{\pi}{2} - \dfrac{\pi}{4}- x\right)= \dfrac{3\sqrt2}{2}$
$\Leftrightarrow 2\sin\left(x +\dfrac{\pi}{4}\right) + \cos\left(x + \dfrac{\pi}{4}\right)= \dfrac{3\sqrt2}{2}$
$\Leftrightarrow \dfrac{2}{\sqrt5}\sin\left(x +\dfrac{\pi}{4}\right) + \dfrac{1}{\sqrt5}\cos\left(x + \dfrac{\pi}{4}\right)= \dfrac{3\sqrt{10}}{10}$
Đặt $\begin{cases}\cos\alpha = \dfrac{2}{\sqrt5}\\\sin\alpha = \dfrac{1}{\sqrt5}\end{cases}\Rightarrow \alpha = \arccos\dfrac{2}{\sqrt5}$
Phương trình trở thành:
$\quad \sin\left(x +\dfrac{\pi}{4}\right)\cos\alpha + \cos\left(x +\dfrac{\pi}{4}\right)\sin\alpha = \dfrac{3\sqrt{10}}{10}$
$\Leftrightarrow \sin\left(x + \dfrac{\pi}{4} + \alpha\right)= \dfrac{3\sqrt{10}}{10}$
$\Leftrightarrow \left[\begin{array}{l}x + \dfrac{\pi}{4} + \alpha = \arcsin\left(\dfrac{3\sqrt{10}}{10}\right) + k2\pi\\x + \dfrac{\pi}{4} + \alpha =\pi - \arcsin\left(\dfrac{3\sqrt{10}}{10}\right) + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = - \dfrac{\pi}{4} - \alpha + \arcsin\left(\dfrac{3\sqrt{10}}{10}\right) + k2\pi\\x = \dfrac{3\pi}{4} -\alpha - \arcsin\left(\dfrac{3\sqrt{10}}{10}\right) + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = - \dfrac{\pi}{4} - \arccos\dfrac{2}{\sqrt5} + \arcsin\left(\dfrac{3\sqrt{10}}{10}\right) + k2\pi\\x = \dfrac{3\pi}{4} -\arccos\dfrac{2}{\sqrt5} - \arcsin\left(\dfrac{3\sqrt{10}}{10}\right) + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$
$c)\quad \sin2x + \sin^2x = \dfrac12$
$\Leftrightarrow \sin2x + \dfrac{1 - \cos2x}{2}=\dfrac12$
$\Leftrightarrow 2\sin2x - \cos2x = 0$
$\Leftrightarrow \dfrac{2}{\sqrt5}\sin2x - \dfrac{1}{\sqrt5}\cos2x = 0$
Đặt $\begin{cases}\cos\alpha = \dfrac{2}{\sqrt5}\\\sin\alpha = \dfrac{1}{\sqrt5}\end{cases}\Rightarrow \alpha = \arccos\dfrac{2}{\sqrt5}$
Phương trình trở thành:
$\quad \sin2x\cos\alpha - \cos2x\sin\alpha = 0$
$\Leftrightarrow \sin(2x -\alpha)= 0$
$\Leftrightarrow 2x - \alpha = k\pi$
$\Leftrightarrow 2x = \alpha + k\pi$
$\Leftrightarrow x = \dfrac12\alpha + \dfrac{k\pi}{2}$
$\Leftrightarrow x = \dfrac12\arccos\dfrac{2}{\sqrt5} + \dfrac{k\pi}{2}\quad (k\in\Bbb Z)$
