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

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

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2cos^2(θ)+3cos(θ)-2=0

2 cos^{2} (θ) + 3 cos (θ) - 2 = 0

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

Solution steps

2 cos^{2} (θ) + 3 cos (θ) - 2 = 0

Solve by substitution

cos (θ) = \frac{1}{2}, cos (θ) = - 2

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

cos (θ) = - 2 :

No Solution

Combine all the solutions

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

12 (1 - cos (θ)) = sin^{2} (θ)

sin (\frac{1}{6} x) = 0

2 sin (x - \frac{π}{6}) = cos (x - \frac{π}{3})

- 3 sin^{2} (θ) + 5 sin (θ) - 1 = 1

14 sin^{2} (x) + 21 sin (x) + 7 = 0

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