Question

The number of real solutions of the equation $2\sin 3x+\sin 7x-3=0$ which lie in the interval $[-2\pi ,2\pi ]$ is?

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

$2\text{sin}3x+\text{sin}7x=3$

For this to be true, both $\text{sin}3x$ and $\text{sin}7x$ must be equal to $1$

Now,

For $\text{sin}3x=1$

$\Rightarrow x=\frac{\pi }{6},\frac{5\pi }{6},\frac{9\pi }{6},\frac{-\pi }{2},\frac{-7\pi }{6},\frac{-11\pi }{6}$

For $\text{sin}7x=1$

$\Rightarrow x=\frac{(4n+1)\pi }{14}$ for $n=0,1,2,3,4,5,6$

$=-\frac{(4n-1)\pi }{14}$ for $n=1,2,3,4,5,6,7$

So, only possible values of $x$ are $\frac{3\pi }{2}$ and $\frac{-\pi }{2}$