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

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

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sin^2(θ)-sin(θ)+1=cos^2(θ)

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

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

Solution steps

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

Subtract

cos^{2} (θ)

from both sides

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

Rewrite using trig identities

- sin (θ) + 2 sin^{2} (θ) = 0

Solve by substitution

sin (θ) = \frac{1}{2}, sin (θ) = 0

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

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

Combine all the solutions

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

1 = l cos (x)

tan (2 x) = cos (2 x), 0 \leq x \leq π

5 sin (2 x) = 5 cos (x), 0 \leq x \leq 2 π

x, f = cos (x) cos (h y)

tan (x) = - \frac{1}{10}

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