Answer:
LHS=2(sin
6
θ+cos
θ)−3(sin
4
θ)+1
=2{(sin
2
θ)
3
−3sin
θcos
θ(sin
θ)}−3(sin
−2(sin
θ)}+1
We know, [sin²x+cos²x=1]
=2{1−3sin
θ}−3{1−2sin
θ}+1
=2−6sin
θ−3+6sin
θ+1
=0
=RHS
Explanation:
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