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

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

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cos^2(θ)-sin^2(θ)=-sin(θ)+1

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

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) = - sin (θ) + 1

Subtract

- sin (θ) + 1

from both sides

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

2 sin (θ) + 6 = 7

8 cos (12 x) + 4 = - 4

12 tan^{3} (x) = 4 tan (x)

cos (2 θ) = 3 sin (θ) - 1

sin (4 x) = - 1

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