Question

The solution set of the equation $\sin 3x+\cos 2x=-2$ is
(where $n\in Z$)

• A. $\frac{(2n+1)\pi }{2}$
• B. $\frac{(4n-1)\pi }{2}$
• C. $\frac{(2n-1)\pi }{2}$
• D. $\frac{(4n+1)\pi }{2}$

Answer 1

The correct option is D $\frac{(4n+1)\pi }{2}$

Since $\sin 3x\ge -1\text{and}\cos 2x\ge -1$
we have
$\sin 3x+\cos 2x\ge -2$
Thus the equality holds true if and only if;
$\sin 3x=-1\text{and}\cos 2x=-1$
Now,
$\sin 3x=-1$
$\Rightarrow 3x=\frac{(4n-1)\pi }{2},n\in Z\Rightarrow x=\frac{(4n-1)\pi }{6}\Rightarrow x=-\frac{\pi }{6},\frac{3\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6},\frac{15\pi }{6},\cdots \Rightarrow x=-\frac{\pi }{6},\frac{\pi }{2},\frac{7\pi }{6},\frac{11\pi }{6},\frac{5\pi }{2},\cdots$
$\cos 2x=-1\Rightarrow 2x=(2n+1)\pi \Rightarrow x=\frac{(2n+1)\pi }{2},n\in Z$
$\Rightarrow x=\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},\cdots$
Therefore the requires solution set is
$x=\frac{\pi }{2},\frac{5\pi }{2},\frac{9\pi }{2}\therefore x=\frac{(4n+1)\pi }{2},n\in Z$