If cos theta cos3q cos2q 1 4 where 0 le theta le pi find the value of q that satisfi...........

Mathematics (NDA) Trigonometry

. If cosθ.cos3q.cos2q =1/4 where 0≤θ≤ π, find the value of q that satisfies the given equation?

θ =kπ±π/2 , (2n +1)π/8

θ =kπ±π/3 , (2n +1)π/4

θ =kπ±π/3 , (2n +1)π/8

θ =kπ±π/3 , (2n +1)π/3

cosθ cos3q cos2q =1/4

(cos2θ +cos 4θ)cos2θ =1/4

½(2 cos² 2θ +2 cos 4θ cos2θ)=1/2

1+ cos 4θ +2 cos 4θ cos2θ =1

cos 4θ (1+ 2 cos2θ )=0

cos 4θ =0 or 1+2cos 2θ =0

cos 4θ =0

cos 4θ =cosπ/2

4θ=(n+1/2)π

Θ=(2n +1)π/8

1+2cos 2θ =0

cos 2θ =-1/2

cos 2θ =-cos π/3

θ =kπ±π/3

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