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Question

The number of solutions of 4cos2(π4x2)+4sin4x+sin22x=0 in x[0,π] is

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Solution

4cos2(π4x2)+4sin4x+sin22x=0
Since, cos2(π4x2)0 and 4sin4x+sin22x0

So, the solution of above equation is exist only when
cos2(π4x2)=0 and 4sin4x+sin22x=0
12[cos(π2x)+1]=0 and 4sin4x+4sin2xcos2x=0
sinx+1=0 and 4sin2x=0sinx=1 and sinx=0
Which is not possible.

Hence, there is no solution.

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