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Question

Number of solutions of the equation
|sinx|=sinx2cosx in [0,4π] is

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Solution

Case 1: sinx0
sin x=sinx2cosxcosx=0
x=π2,3π2,5π2,7π2
Since, sin x0x=π2,5π2

Case 2: sinx<0
sinx=sinx2cosx
2sinx=2cosxtanx=1
x=π4,5π4,9π4,13π4
Since, sinx<0x=5π4,13π4
Number of solutions is 4.

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