What is the general solution for cos^2(θ)-3cos(θ)+2=sin^2(θ) ?

The general solution for cos^2(θ)-3cos(θ)+2=sin^2(θ) is θ=2pin,θ= pi/3+2pin,θ=(5pi)/3+2pin

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cos^2(θ)-3cos(θ)+2=sin^2(θ)

cos^{2} (θ) - 3 cos (θ) + 2 = sin^{2} (θ)

θ = 2 πn, θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Solution steps

cos^{2} (θ) - 3 cos (θ) + 2 = sin^{2} (θ)

Subtract

sin^{2} (θ)

from both sides

cos^{2} (θ) - 3 cos (θ) + 2 - sin^{2} (θ) = 0

Rewrite using trig identities

1 + 2 cos^{2} (θ) - 3 cos (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{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Combine all the solutions

θ = 2 πn, θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

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

2 + 2 sin (θ) = - 4 sin (θ)

cos (θ) = \frac{( 5 )}{( 13 )}

2 sin (θ) + 3 = - csc (θ)

2 sin (\frac{2 x}{3}) - 2 = 0

What is the general solution for cos^2(θ)-3cos(θ)+2=sin^2(θ) ?
The general solution for cos^2(θ)-3cos(θ)+2=sin^2(θ) is θ=2pin,θ= pi/3+2pin,θ=(5pi)/3+2pin