What is the general solution for 2sin^2(θ)=3(cos(θ)+1) ?

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

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2sin^2(θ)=3(cos(θ)+1)

2 sin^{2} (θ) = 3 (cos (θ) + 1)

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

Solution steps

2 sin^{2} (θ) = 3 (cos (θ) + 1)

Subtract

3 (cos (θ) + 1)

from both sides

2 sin^{2} (θ) - 3 (cos (θ) + 1) = 0

Rewrite using trig identities

- 1 - 2 cos^{2} (θ) - 3 cos (θ) = 0

Solve by substitution

cos (θ) = - 1, cos (θ) = - \frac{1}{2}

cos (θ) = - 1 : θ = π + 2 πn

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

Combine all the solutions

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

sin (θ) = 0.3

tan (\frac{x}{2}) + 3 = 0

tan^{2} (θ) - 4 = 0

csc^{2} (θ) - 2 csc (θ) - 8 = 0

(tan (θ) + 1) (cos (θ) + 1) = 0

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