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The number of solutions of the equation

1+\sin x .\sin ^2\frac{x}{2}=0\:\:in\:\:[-\pi,\pi] are

Option: 1

0


Option: 2

1


Option: 3

2


Option: 4

3


Answers (1)

best_answer

We have

        1 + sinx.sin^{2}\frac{x}{2}= 0 

        1 + sinx\left( {\frac{{1 - \cos x}}{2}} \right)=0 

        2 + sinx - sinxcosx = 0

        sinx.cosx = 2 + sinx

Multiplying the equation by 2

        sin2x  = 4 + 2sinx

Now maximum value of LHS = 1 and minimum value of RHS = 4 - 2 = 2

So LHS and RHS can never be equal. Hence the equation has no solution

 

Posted by

vishal kumar

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