Đáp án:

$\left[\begin{array}{l}x = -\dfrac{\pi}{2} + k2\pi\\x = \pi + k2\pi\\x = -\dfrac{\pi}{4}-\arcsin\left(\dfrac{1}{2\sqrt2}\right) + k2\pi\\x =\dfrac{3\pi}{4} + \arcsin\left(\dfrac{1}{2\sqrt2}\right) + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$

Giải thích các bước giải:

$\quad 3(\sin x + \cos x) + 2\sin2x = -3$

Đặt $t = \sin x +\cos x \quad \left(|t|\leqslant \sqrt2\right)$

$\Rightarrow t^2 - 1 = \sin2x$

Phương trình trở thành:

$\quad 3t + 2(t^2 -1)+3= 0$

$\Leftrightarrow 2t^2 + 3t + 1 = 0$

$\Leftrightarrow (t+1)(2t+1)= 0$

$\Leftrightarrow \left[\begin{array}{l}t = -1\\t = -\dfrac12\end{array}\right.$ (nhận)

+) Với $t = -1$ ta được:

$\quad \sin x +\cos x = -1$

$\Leftrightarrow \sqrt2\sin\left(x +\dfrac{\pi}{4}\right)= -1$

$\Leftrightarrow \sin\left(x +\dfrac{\pi}{4}\right)=-\dfrac{\sqrt2}{2}$

$\Leftrightarrow \left[\begin{array}{l}x +\dfrac{\pi}{4} = -\dfrac{\pi}{4} + k2\pi\\x + \dfrac{\pi}{4} = \pi +\dfrac{\pi}{4} + k2\pi\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l}x = -\dfrac{\pi}{2} + k2\pi\\x = \pi + k2\pi\end{array}\right.$

+) Với $t= -\dfrac12$ ta được:

$\quad \sin x + \cos x = -\dfrac12$

$\Leftrightarrow \sqrt2\sin\left(x +\dfrac{\pi}{4}\right)= -\dfrac12$

$\Leftrightarrow \sin\left(x +\dfrac{\pi}{4}\right)= -\dfrac{1}{2\sqrt2}$

$\Leftrightarrow \left[\begin{array}{l}x +\dfrac{\pi}{4}=\arcsin\left(-\dfrac{1}{2\sqrt2}\right) + k2\pi\\x +\dfrac{\pi}{4}=\pi - \arcsin\left(-\dfrac{1}{2\sqrt2}\right) + k2\pi\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l}x = -\dfrac{\pi}{4}-\arcsin\left(\dfrac{1}{2\sqrt2}\right) + k2\pi\\x =\dfrac{3\pi}{4} + \arcsin\left(\dfrac{1}{2\sqrt2}\right) + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$

Vậy phương trình có các họ nghiệm là:

$x =-\dfrac{\pi}{2}+ k2\pi$

$x=\pi+ k2\pi$

$x = -\dfrac{\pi}{4}-\arcsin\left(\dfrac{1}{2\sqrt2}\right) + k2\pi$

$x =\dfrac{3\pi}{4} + \arcsin\left(\dfrac{1}{2\sqrt2}\right) + k2\pi$

với $k\in\Bbb Z$

`~rai~`

\(3(\sin x+\cos x)+2\sin2x=-3\\\text{+)Đặt t=}\sin x+\cos x\quad(-\sqrt{2}\le t\le\sqrt{2})\\\text{+)Ta có:}t^2=(\sin x+\cos x)^2=\sin^2 x+2\sin x\cos x+\cos^2x\\\Leftrightarrow t^2=1+\sin2x\\\Leftrightarrow \sin2x=t^2-1.\\\text{Khi đó,phương trình (1) trở thành:}\\3t+2(t^2-1)=-3\\\Leftrightarrow 2t^2+3t-2+3=0\\\Leftrightarrow 2t^2+3t+1=0\quad(a-b+c=0)\\\Leftrightarrow \left[\begin{array}{I}t=-1\\t=-\dfrac{1}{2}\end{array}\right.\text{(thỏa mãn)}\\\Leftrightarrow \left[\begin{array}{I}\sin x+\cos x=-1\\\sin x+\cos x=-\dfrac{1}{2}\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}\sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right)=-1\\\sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{1}{2}\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}\sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}\\\sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{4}\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x+\dfrac{\pi}{4}=-\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\pi+\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\arcsin\left(-\dfrac{\sqrt{2}}{4}\right)+k2\pi\\x+\dfrac{\pi}{4}=\pi-\arcsin\left(-\dfrac{\sqrt{2}}{4}\right)+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=-\dfrac{\pi}{2}+k2\pi\\x=\pi+k2\pi\\x=\arcsin\left(-\dfrac{\sqrt{2}}{4}\right)-\dfrac{\pi}{4}+k2\pi\\x=-\arcsin\left(-\dfrac{\sqrt{2}}{2}\right)+\dfrac{3\pi}{4}+k2\pi\end{array}\right.\quad(k\in\mathbb{Z})\)