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Q. The Number of solutions of the equation $\sin ^{3} x \cos x+\sin ^{2} x\cos ^{2} x+\sin x \cos ^{3} x=1$ in the interval $[0,2 \pi]$ is

Trigonometric Functions

Solution:

$\sin ^{3} x \cos x+\sin ^{2} x \cos ^{2} x+\sin x \cos ^{3} x=1$
$\Rightarrow \sin x \cos x\left(\sin ^{2} x+\sin x \cos x+\cos ^{2} x\right)=1$
$\Rightarrow \frac{\sin 2 x}{2}\left(1+\frac{\sin 2 x}{2}\right)=1$
$\Rightarrow \sin 2 x(2+\sin 2 x)=4$
$\Rightarrow \sin ^{2} 2 x+2 \sin 2 x-4=0$
$\Rightarrow \sin 2 x=\frac{-2 \pm \sqrt{4+16}}{2}$
$\Rightarrow \sin 2 x=-1 \pm \sqrt{5}$
This is not possible. (since $-1 \leq \sin 2 x \leq 1$ )
Hence the given equation has no solution.