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

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

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-sin^2(θ)+cos^2(θ)=1-sin(θ)

- sin^{2} (θ) + cos^{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

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

Subtract

1 - sin (θ)

from both sides

- sin^{2} (θ) + cos^{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

1 - sin (θ) = 2 cos^{2} (θ)

2 sin^{2} (θ) - 3 sin (θ) = 0

sec^{2} (x) + 2 tan (x) = 4

tan (\frac{x}{5}) + 3 = 0

cos (2 x) - 2 cos (x) = - 1

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