The number of real solutions of the equation `sqrt(1+cos2x)=sqrt(2)sin^(-1)(sinx),-pilt=xlt=pi` is 0 (b) 1 (c) 2 (d) infinite

Explanation

We have given

`sqrt(1+cos2x)=sqrt(2)sin^(-1)(sinx)`

`sqrt(1+2cos^2x-1)=sqrt(2)sin^(-1)(sinx)`

`sqrt(2cos^2x)=sqrt(2)sin^(-1)(sinx)`

Now for `-(pi)/2lt=xlt=(pi)/2`

`sqrt(2cos^2x)=sqrt(2)sin^(-1)(sinx)`

`sqrt(2)cosx=sqrt(2)x`

`cosx=x`

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