Question
Mathematics Question on Trigonometric Functions
If sin6θ+sin4θ+sin2θ=0, then general value of θ is
A
4nπ,nπ±3π
B
4nπ,nπ±4π
C
4nπ,nπ±6π
D
4nπ,2nπ±6π
Answer
4nπ,nπ±3π
Explanation
Solution
sin6θ+sin4θ+sin2θ=0 ⇒(sin6θ+sin2θ)+sin4θ=0 ⇒2sin4θcos2θ+sin4θ=0 ⇒sin4θ(2cos2θ+1)=0 ⇒sin4θ=0 or cos2θ=−21=cos32π ⇒4θ=nπ or 2θ=2nπ±32 ⇒θ=4nπ,θ=nπ±3πθ