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  • The number of solutions of the equation <img alt="1+\sin x \sin ^2\left(\frac{x}{2}\right)=0 in [-\pi, \pi]" src="https://entrancecorner.oncodecogs.com/gif.latex?1&plus;%5Csin%20x%20%5Csin%20%5

The number of solutions of the equation 1+\sin x \sin ^2\left(\frac{x}{2}\right)=0 in [-\pi, \pi]is

 

Option: 1

0


Option: 2

1


Option: 3

2


Option: 4

3


Answers (1)

best_answer

\begin{aligned} & 1+\sin x \sin ^2 \frac{x}{2}=0 \\ & 1+\sin x\left(\frac{1-\cos x}{2}\right)=0 \end{aligned}
\begin{array}{cc} \Rightarrow & 2+\sin x-\sin x \cos x=0 \\ \Rightarrow & 4+2 \sin x=\sin 2 x \end{array}
Hence, no solution ie, Number of solutions $=$ zero Alternative Method : Given equation is

\sin x \sin ^2\left(\frac{x}{2}\right)=-1

\text { It could be possible only, if }
\begin{aligned} & & \quad \sin x=-1 \text { and } \sin ^2\left(\frac{x}{2}\right)=1 \\ & (\because-1 \leq \sin x \leq 1) \end{aligned}

 Which is not possible simultaneously.

Posted by

Devendra Khairwa

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