What is the general solution for 2cos^2(θ)+3sin(θ)-3=0 ?

The general solution for 2cos^2(θ)+3sin(θ)-3=0 is θ= pi/6+2pin,θ=(5pi)/6+2pin,θ= pi/2+2pin

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2cos^2(θ)+3sin(θ)-3=0

2 cos^{2} (θ) + 3 sin (θ) - 3 = 0

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

Solution steps

2 cos^{2} (θ) + 3 sin (θ) - 3 = 0

Rewrite using trig identities

- 1 - 2 sin^{2} (θ) + 3 sin (θ) = 0

Solve by substitution

sin (θ) = \frac{1}{2}, sin (θ) = 1

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

sin (θ) = 1 : θ = \frac{π}{2} + 2 πn

Combine all the solutions

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

sec (θ) = \frac{4}{5}

- 2 = 22 cos (θ)

cos^{2} (2 x) = 1

- 4 cos^{2} (θ) = - 1

- sin (2 x) + sin (x) = 0

What is the general solution for 2cos^2(θ)+3sin(θ)-3=0 ?
The general solution for 2cos^2(θ)+3sin(θ)-3=0 is θ= pi/6+2pin,θ=(5pi)/6+2pin,θ= pi/2+2pin