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

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

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-3cos(θ)+3=2sin^2(θ)

- 3 cos (θ) + 3 = 2 sin^{2} (θ)

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

Solution steps

- 3 cos (θ) + 3 = 2 sin^{2} (θ)

Subtract

2 sin^{2} (θ)

from both sides

- 3 cos (θ) + 3 - 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

2 = 2 sin (θ)

sin (a) = 1

4 cos (2 t) = 0

cos (3 x) = - \frac{3}{2}

cos (x) = \frac{2}{7}

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