Question

Number of roots of the equation $\sin x+2\sin 2x=3+\sin 3x$ in $[0,\pi ]$ is/are

• A. 2
• B. 3
• C. 0
• D. 4

Answer 1

The correct option is C 0

$\sin x+2\sin 2x=3+\sin 3x$

$\Rightarrow \sin x+4\sin x.\cos x=3+(3\sin x-4{\sin }^{3}x)$

$\Rightarrow 4{\sin }^{3}x-2+4\sin x.\cos x=3$

$\Rightarrow \sin x(4{\sin }^2+4\cos x-2)=3$

$\Rightarrow \sin x[3-(4{\cos }^{2}x-4\cos x+1\left)\right]=3$

$\Rightarrow \sin x[3-(2\cos x-1{)}^2]=3$

Now $\sin x[3-(2\cos x-1{)}^2]<3$

Thus there won't exist any solution.
Hence, option 'C' is correct.