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

The general solution for 2sin^2(θ)-3cos(θ)=3 is θ=pi+2pin,θ=(2pi)/3+2pin,θ=(4pi)/3+2pin

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2sin^2(θ)-3cos(θ)=3

2 sin^{2} (θ) - 3 cos (θ) = 3

θ = π + 2 πn, θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

Solution steps

2 sin^{2} (θ) - 3 cos (θ) = 3

Subtract

3

from both sides

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

cos (θ) = - \frac{1}{2} : θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

Combine all the solutions

θ = π + 2 πn, θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

cot (θ) = \frac{1}{3}

sec (x) csc (x) = 2 sec (x)

2 tan^{2} (x) + sec (x) = 1

(tan (θ) - 1) (cos (θ) - 1) = 0

cos (x) = \frac{16}{20}

What is the general solution for 2sin^2(θ)-3cos(θ)=3 ?
The general solution for 2sin^2(θ)-3cos(θ)=3 is θ=pi+2pin,θ=(2pi)/3+2pin,θ=(4pi)/3+2pin