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