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

The general solution for cos^2(θ)-sin^2(θ)=1-sin(θ) is θ=2pin,θ=pi+2pin,θ= pi/6+2pin,θ=(5pi)/6+2pin

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cos^2(θ)-sin^2(θ)=1-sin(θ)

cos^{2} (θ) - sin^{2} (θ) = 1 - sin (θ)

θ = 2 πn, θ = π + 2 πn, θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Solution steps

cos^{2} (θ) - sin^{2} (θ) = 1 - sin (θ)

Subtract

1 - sin (θ)

from both sides

cos^{2} (θ) - sin^{2} (θ) - 1 + sin (θ) = 0

Rewrite using trig identities

sin (θ) - 2 sin^{2} (θ) = 0

Solve by substitution

sin (θ) = 0, sin (θ) = \frac{1}{2}

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin (θ) = \frac{1}{2} : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn, θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

sin (x) cos (x) = \frac{2}{4}

cos^{2} (x) = 1 + sin (x)

3 sec (θ) + 2 = 0

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

cos (θ) = sin (θ)

What is the general solution for cos^2(θ)-sin^2(θ)=1-sin(θ) ?
The general solution for cos^2(θ)-sin^2(θ)=1-sin(θ) is θ=2pin,θ=pi+2pin,θ= pi/6+2pin,θ=(5pi)/6+2pin