Question

The sum of roots of the equation $si{n}^{4}x-co{s}^{2}xsinx+2si{n}^{2}x+sinx=0$ for $0\le x\le 3\pi$ is equal to

• A. $\frac{3\pi }{4}$
• B. $2\pi$
• C. $4\pi$
• D. $6\pi$

Answer 1

Answer: (D)

$\sin x\left({\sin }^{3}x-{\cos }^{2}x+2\sin x+1\right)=0$
$\sin x\left({\sin }^{3}x+2\sin x+1-{\cos }^{2}x\right)=0$
$\sin x\left({\sin }^{3}x+{\sin }^{2}x+2\sin x\right)=0$
${\sin }^{2}x\left({\sin }^{2}x+\sin x+2\right)=0$
$\Rightarrow \sin x=0$ as ${\sin }^{2}x+\sin x+2\ne 0$
$\Rightarrow x=n\pi ,n\in I$
$x\in [0,3\pi ],x=0,\pi ,2\pi ,3\pi$
$\Rightarrow$ sum of roots $=6\pi$