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

The general solution for 2sin^2(θ)=3(1-cos(-θ)) is θ= pi/3+2pin,θ=(5pi)/3+2pin,θ=2pin

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2sin^2(θ)=3(1-cos(-θ))

2 sin^{2} (θ) = 3 (1 - cos (- θ))

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn, θ = 2 πn

Solution steps

2 sin^{2} (θ) = 3 (1 - cos (- θ))

Rewrite using trig identities

2 sin^{2} (θ) = 3 (1 - cos (θ))

Subtract

3 (1 - cos (θ))

from both sides

2 sin^{2} (θ) - 3 (1 - cos (θ)) = 0

Rewrite using trig identities

- 1 - 2 cos^{2} (θ) + 3 cos (θ) = 0

Solve by substitution

cos (θ) = \frac{1}{2}, cos (θ) = 1

cos (θ) = \frac{1}{2} : θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

cos (θ) = 1 : θ = 2 πn

Combine all the solutions

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn, θ = 2 πn

sec (x) - 2 = 0

2 sin^{4} (x) + sin^{2} (x) - 1 = 0

tan (\frac{x}{9}) + 3 = 0

cos (θ) = \frac{6}{9}

sec (b) = 2

What is the general solution for 2sin^2(θ)=3(1-cos(-θ)) ?
The general solution for 2sin^2(θ)=3(1-cos(-θ)) is θ= pi/3+2pin,θ=(5pi)/3+2pin,θ=2pin