Question

The number of solution of the equation, ${\sin }^{4}x-{\cos }^{2}x.\sin x+2{\sin }^{2}x+\sin x=0,0\le x\le 3\pi$, is _________.

Answer 1

$\sin {}^{4}x-\cos {}^{2}x\sin x+2\sin {}^{2}x+\sin x=0\sin {}^{4}x-(1-\sin {}^{2}x)\sin x+2\sin {}^{2}x+\sin x=0{\sin }^{4}x-\sin x+{\sin }^{3}x+2{\sin }^{2}x+\sin x=0{\sin }^{2}x({\sin }^{2}x+\sin x+2)=0$

Since ${\sin }^{2}x+\sin x+2\ne 0$ for any value of x

we can say that ${\sin }^{2}x=0\Rightarrow \sin x=0$

for $0\le x\le 3\pi$

$\sin x=0xϵ\{0,\pi ,2\pi ,3\pi \}$

Thus $4$ solutions.