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Popular >

2sin(2θ)cos(θ)+cos(2θ)sin(θ)=0

Solution

2 sin (2 θ) cos (θ) + cos (2 θ) sin (θ) = 0

Solution

θ = 2 πn, θ = π + 2 πn, θ = - 1.15026 \dots + πn, θ = 1.15026 \dots + πn

+1

Solution steps

2 sin (2 θ) cos (θ) + cos (2 θ) sin (θ) = 0

Rewrite using trig identities

cos (2 θ) sin (θ) + 4 cos^{2} (θ) sin (θ) = 0

Factor

cos (2 θ) sin (θ) + 4 cos^{2} (θ) sin (θ) : sin (θ) (cos (2 θ) + 4 cos^{2} (θ))

sin (θ) (cos (2 θ) + 4 cos^{2} (θ)) = 0

Solving each part separately

sin (θ) = 0 or cos (2 θ) + 4 cos^{2} (θ) = 0

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

cos (2 θ) + 4 cos^{2} (θ) = 0 : θ = arctan (- 5) + πn, θ = arctan (5) + πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn, θ = arctan (- 5) + πn, θ = arctan (5) + πn

Show solutions in decimal form

θ = 2 πn, θ = π + 2 πn, θ = - 1.15026 \dots + πn, θ = 1.15026 \dots + πn

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