Solve the equation 3 – 2cos θ – 4 sin θ – cos 2 θ + sin 2 θ = 0
Text Solution
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= (4n + 1) π /2, θ = 2n π , n ∈ Ι
3 – 2cos θ – 4 sin θ – cos 2 θ + sin 2 θ = 0
⇒ 3 – 2cos θ – 4 sin θ – 2cos 2 θ + 1 + 2sin θ cos θ = 0
⇒ 4 – 2cos θ – 4 sin θ – 2cos 2 θ + 2sin θ cos θ = 0
⇒ 2 – 2cos θ – 4 sin θ + 2sin2 θ + 2sin θ cos θ = 0
⇒ 2 – 2(cos θ +sin θ ) –2sin θ +2sin θ (sin θ +cos θ ) = 0
⇒ (sin θ + cos θ ) (– 2 + 2sin θ ) + 2 – 2sin θ = 0 ⇒ (sin θ + cos θ –1) (– 2 + 2sin θ ) = 0
⇒ sin θ = 1 or sin θ + cos θ = 1 ⇒ 
or 
= 
⇒ 
or 
= 
⇒ 
, 2n π
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