The number of solutions of the equation `sin^3xcosx+sin^2xcos^2x+sinxcos^3x=1` in the interval `[0,2pi]` is/are

Options

A. 0
B. 2
C. 3
D. infinite

Explanation

`sin^3xcosx+sin^2xcos^2x+sinxcos^3x=1`

`sinxcosx(sin^2x+sinxcosx+cos^2x)=1`

`sinxcosx(1+sinxcosx)=1`

`(sin2x)/2*[1+(sin2x)/2]=1`

`sin^2 2x+2sinx-4=0`

`sin2x=(-2pmsqrt(4+16))/2`

`=(-2pmsqrt20)/2=-1pmsqrt5`

`sin2x=-1+2.326=1.326>=1`

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