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

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

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-3sin(θ)+3=2cos^2(θ)

- 3 sin (θ) + 3 = 2 cos^{2} (θ)

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

Solution steps

- 3 sin (θ) + 3 = 2 cos^{2} (θ)

Subtract

2 cos^{2} (θ)

from both sides

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

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

Combine all the solutions

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

2 sin (x - \frac{π}{2}) + 1 = 0

2 cos^{2} (x) - 1 = cos (x)

0 = 2 tan (x) - sec^{2} (x)

csc^{2} (x) + 6 csc (x) - 7 = 0

\frac{sin ( 2 x ) - 1}{3} = - \frac{1}{6}

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