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

The general solution for 2cos^2(θ)=3sin(θ)+3 is θ=(3pi)/2+2pin,θ=(7pi)/6+2pin,θ=(11pi)/6+2pin

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2cos^2(θ)=3sin(θ)+3

2 cos^{2} (θ) = 3 sin (θ) + 3

θ = \frac{3 π}{2} + 2 πn, θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Solution steps

2 cos^{2} (θ) = 3 sin (θ) + 3

Subtract

3 sin (θ) + 3

from both sides

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) = - 1, sin (θ) = - \frac{1}{2}

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

sin (θ) = - \frac{1}{2} : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Combine all the solutions

θ = \frac{3 π}{2} + 2 πn, θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

cos (π x^{2}) = 0

sin^{2} (θ) + sin (θ) = 0, 0 \leq θ < 2 π

16 tan^{2} (θ) - 1 = 0

sin^{2} (x) + \frac{5}{14} = \frac{6}{7}

sin (2 θ) - 2 sin (θ) = 0

What is the general solution for 2cos^2(θ)=3sin(θ)+3 ?
The general solution for 2cos^2(θ)=3sin(θ)+3 is θ=(3pi)/2+2pin,θ=(7pi)/6+2pin,θ=(11pi)/6+2pin