Question

# For the given equation, find the number of solutions in has,$\sin x + 2\sin 2x - \sin 3x = 3,x \in \left( {0,\pi } \right)$(A) Infinitely many solutions (B) three solutions (C) one solution (D) no solution

Hint: Use $\sin 2x = 2\sin x\cos x$ and $\sin 3x = 3\sin x - 4{\sin ^3}x$ and then simplify the equation.
According to question, the given equation is:
$\sin x + 2\sin 2x - \sin 3x = 3$
We know that, $\sin 2x = 2\sin x\cos x$ and $\sin 3x = 3\sin x - 4{\sin ^3}x$, using these two results, we’ll get:
$\Rightarrow \sin x + 2(2\sin x\cos x) - (3\sin x - 4{\sin ^3}x) = 3 \\ \Rightarrow \sin x + 4\sin x\cos x - 3\sin x + 4{\sin ^3}x = 3 \\ \Rightarrow \sin x[1 + 4\cos x - 3 + 4{\sin ^2}x] = 3 \\ \Rightarrow \sin x[4\cos x + 4{\sin ^2}x - 2] = 3 \\$
Now, putting ${\sin ^2}x = 1 - {\cos ^2}x$ we’ll get:
$\Rightarrow \sin x[4\cos x + 4 - 4{\cos ^2}x - 2] = 3 \\ \Rightarrow \sin x[2 - (4{\cos ^2}x - 4\cos x)] = 3 \\ \Rightarrow \sin x[2 - (4{\cos ^2}x - 4\cos x + 1) + 1] = 3 \\ \Rightarrow \sin x[3 - {(2\cos x - 1)^2}] = 3 \\$
We know that, $3 - {(2\cos x - 1)^2} \geqslant 3$ Therefore, for the above equation to satisfy, we have:
$\Rightarrow \sin x = 1$ and ${(2\cos x - 1)^2} = 0$
$\Rightarrow x = \frac{\pi }{2}$ and $\cos x = \frac{1}{2}$
$\Rightarrow x = \frac{\pi }{2}$ and $x = \frac{\pi }{3}$
But $x$ cannot have two values at the same time. Therefore, the above equation will not have any solution. And option (D) is correct.

Note: Both $\sin x = 1$ and $\cos x = \frac{1}{2}$ cannot satisfy at the same time. If in any equation, the value of $\sin x$ is coming out to be $1,$ for any value of $x,$ then the value of $\cos x$ must be $0$ for that particular $x.$