If sin6θ=32cos5θsinθ-32cos3θsinθ+3x, then x=
cosθ
cos2θ
sinθ
sin2θ
Explanation for the correct option.
Given that, sin6θ=32cos5θsinθ-32cos3θsinθ+3x....(1)
We know that, sin2θ=2sinθcosθ consider the LHS
2sin3θcos3θ=23sinθ-4sin3θ4cos3θ-3cosθ=212sinθcos3θ-9sinθcosθ-16sin3θcos3θ+12sin3θcosθ=24sinθcos3θ-18sinθcosθ-32sin3θcos3θ+24sin3θcosθ=24sinθcosθ1-sin2θ-18sinθcosθ-32sinθ1-cos2θcos3θ+24sin3θcosθ=24sinθcosθ-24sin3θcosθ-18sinθcosθ-32sinθcos3θ+32cos5θsinθ+24sin3cosθ=6sinθcosθ-32sinθcos3θ+32cos5θsinθ=32cos5θsinθ-32cos3θsinθ+3sin2θ....(2)
Comparing (1) & (2) we have:
x=sin2θ
Hence, option D is correct.