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

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

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-2sin^2(θ)+1=cos(θ)

- 2 sin^{2} (θ) + 1 = cos (θ)

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

Solution steps

- 2 sin^{2} (θ) + 1 = cos (θ)

Subtract

cos (θ)

from both sides

- 2 sin^{2} (θ) + 1 - cos (θ) = 0

Rewrite using trig identities

- 1 - cos (θ) + 2 cos^{2} (θ) = 0

Solve by substitution

cos (θ) = 1, cos (θ) = - \frac{1}{2}

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

x, sin (x y) = y

12 arcsin (x) = 2 π

csc (x) = - 1 - cot (x)

cos (2 θ) + 5 cos (θ) + 3 = 0

2 cos (x) = 2

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