Question

If sin 6θ + sin 4θ + sin 2θ = 0, then the general value of θ is

(a) \(\frac{nπ}{4}\), nπ ±  \(\frac{π}{3}\) 

(b)  \(\frac{nπ}{4}\), nπ ±  \(\frac{π}{6}\) 

(c)  \(\frac{nπ}{4}\), 2nπ ±  \(\frac{π}{3}\) 

(d)  \(\frac{nπ}{4}\), 2nπ ±  \(\frac{π}{6}\) 

Answer 1

Answer : (a) \(\frac{n\pi}{4}\),  nπ ± \(\frac{\pi}{3}\) 

sin 6θ + sin 4θ + sin 2θ = 0 

⇒ (sin 6θ + sin 2θ) + sin 4θ = 0 

⇒ 2 sin 4θ cos 2θ + sin 4θ = 0 

⇒ sin 4θ (2 cos 2θ + 1) = 0 

⇒ Either sin 4θ = 0 or (2 cos 2θ + 1) = 0

∴ θ = \(\frac{n\pi}{4}\) or nπ ± \(\frac{\pi}{3}\) , n ∈ I