Question

The number of solution of the equation $\cos 3x+\sin (2x-\frac{7\pi }{6})=-2$ for $[0,2\pi ]$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is B $1$

Given:
$\cos 3x+\sin (2x-\frac{7\pi }{6})=-2$

$\sin \theta \in [-1,1]\cos \theta \in [-1,1]$
So this is possible only when
$\cos 3x=-1\sin (2x-\frac{7\pi }{6})=-1$

$\cos 3x=-1$
$\Rightarrow 3x=(2n+1)\pi ,n\in Z$
$\Rightarrow x=(2n+1)\frac{\pi }{3}\cdots \left(1\right)$

$\sin (2x-\frac{7\pi }{6})=-1$
$\Rightarrow (2x-\frac{7\pi }{6})=(4n-1)\frac{\pi }{2},n\in Z\Rightarrow 2x=\frac{7\pi }{6}-\frac{\pi }{2}+2n\pi \Rightarrow x=n\pi +\frac{\pi }{3}\Rightarrow x=\frac{(3n+1)\pi }{3}\cdots \left(2\right)$

Solutions in $[0,2\pi ]$
From equation $\left(1\right)$,
$x=\frac{\pi }{3},\pi ,\frac{5\pi }{3}$
From equation $\left(2\right)$,
$x=\frac{\pi }{3},\frac{4\pi }{3}$

Hence, the required number of solution is $1$.