(i) sin θ + sin 3 θ + sin 5 θ = 0 ⇒ sin θ + sin 5 θ + sin 3 θ = 0

2 sin 3 θ cos 2 θ + sin 3 θ = 0 ⇒ sin 3 θ = 0 and cos 2 θ = –

θ = , n ∈ Ι , θ = π , n ∈ Ι

(ii) cos θ + sin θ = cos 2 θ + sin 2 θ⇒ cos θ – cos 2 θ = sin 2 θ – sin θ

⇒ 2 sin sin = 2 cos sin ⇒ sin = 0 or tan = 1

⇒ = n π or = n π + ⇒θ = 2n π or θ = +

(iii) cos 2 x + cos 2 2x + cos 2 3x = 1 ⇒ + + = 1

⇒ cos2x + cos4x + cos6x = – 1 ⇒ 2cos4x cos2x = –2cos 2 2x

⇒ cos2x = 0 or cos4x + cos2x = 0

⇒ 2x = (2n +1) or 2cos3x cosx = 0

⇒ , (2n + 1) , (2n + 1)

Now x = (2n + 1) = + may also be written as

x = (3k + 1) + ,(3k + 2) + , (3k) +

= , , = (k + 1) ,

( is same as (2n + 1) ) =

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