Find all the values of θ satisfying the equation : sinθ+sin5θ=sin3θ such that 0≤θ≤π.
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Solution
sinθ+sin5θ=sin3θ⇒sinθ−sin3θ+sin5θ=0⇒3sinθ−16sin3θ+16sin5θ=0⇒sinθ(2sinθ−1)(1+2sinθ)(4sin2θ−3)=0 ⇒sinθ=0 or sinθ=12 or sinθ=−12 or sinθ=±√32 For 0≤θ≤π We get θ=0,π6,π3,2π3,5π6,π