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Question

Number of solutions of the equation |sin4θ|=1 in θ[0,2π] is 


Solution

|sin4θ|=1
sin4θ=±1

When sin4θ=1
4θ=(4n+1)π2, nZ
θ=(4n+1)π8, nZ
Number of solutions in [0,2π] are 4.
i.e. {π8,5π8,9π8,13π8}

When sin4θ=1
4θ=(4n1)π2, nZ
θ=(4n1)π8, nZ
Number of solutions in [0,2π] are 4.
i.e. {3π8,7π8,11π8,15π8}

Hence, total number of solutions =8

Alternate Solution:
We know that |sinθ| is periodic with period π.
So, |sin4θ| will be periodic with period π4.
Graph of y=|sin4θ| in θ[0,π] is


So, number of solution in [0,π/4] is 1.
Number of solution in [0,2π] is 8.

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