What is the general solution for sin(2θ)+sin(θ)+2cos(θ)+1=0 ?

The general solution for sin(2θ)+sin(θ)+2cos(θ)+1=0 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin,θ=(3pi)/2+2pin

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sin(2θ)+sin(θ)+2cos(θ)+1=0

sin (2 θ) + sin (θ) + 2 cos (θ) + 1 = 0

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Solution steps

sin (2 θ) + sin (θ) + 2 cos (θ) + 1 = 0

Rewrite using trig identities

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

Factor

1 + sin (θ) + 2 cos (θ) + 2 cos (θ) sin (θ) : (1 + 2 cos (θ)) (sin (θ) + 1)

(1 + 2 cos (θ)) (sin (θ) + 1) = 0

Solving each part separately

1 + 2 cos (θ) = 0 or sin (θ) + 1 = 0

1 + 2 cos (θ) = 0 : θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

sin (θ) + 1 = 0 : θ = \frac{3 π}{2} + 2 πn

Combine all the solutions

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn, θ = \frac{3 π}{2} + 2 πn

tan (θ) = \frac{563}{2000}

cot (\frac{x}{3}) = 1

tan (t) = 7

cos (x) + cos (x) = 0

3 cos (x) = 2 + 2 sin (x)

What is the general solution for sin(2θ)+sin(θ)+2cos(θ)+1=0 ?
The general solution for sin(2θ)+sin(θ)+2cos(θ)+1=0 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin,θ=(3pi)/2+2pin