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

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

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3cos(θ)+3=2sin^2(θ)

3 cos (θ) + 3 = 2 sin^{2} (θ)

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

Solution steps

3 cos (θ) + 3 = 2 sin^{2} (θ)

Subtract

2 sin^{2} (θ)

from both sides

3 cos (θ) + 3 - 2 sin^{2} (θ) = 0

Rewrite using trig identities

1 + 2 cos^{2} (θ) + 3 cos (θ) = 0

Solve by substitution

cos (θ) = - \frac{1}{2}, cos (θ) = - 1

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

cos (θ) = - 1 : θ = π + 2 πn

Combine all the solutions

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

- 33 + 7 csc (\frac{θ}{5} + 2 3^{\circ}) = 53

10 sin (x) tan (x) + tan (x) = 0

2 sin^{2} (θ) + sin (θ) - 1 = 0, 0 \leq θ \leq 2 π

10 sin (x) cos (x) - 7 cos (x) = 0

sin (x + \frac{π}{4}) - sin (x - \frac{π}{4}) = 1

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