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

The general solution for 2cos(2θ)-2=-3sin(θ) is θ=2pin,θ=pi+2pin,θ=0.84806…+2pin,θ=pi-0.84806…+2pin

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2cos(2θ)-2=-3sin(θ)

2 cos (2 θ) - 2 = - 3 sin (θ)

θ = 2 πn, θ = π + 2 πn, θ = 0.84806 \dots + 2 πn, θ = π - 0.84806 \dots + 2 πn

Solution steps

2 cos (2 θ) - 2 = - 3 sin (θ)

Subtract

- 3 sin (θ)

from both sides

2 cos (2 θ) - 2 + 3 sin (θ) = 0

Rewrite using trig identities

3 sin (θ) - 4 sin^{2} (θ) = 0

Solve by substitution

sin (θ) = 0, sin (θ) = \frac{3}{4}

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

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

Combine all the solutions

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

Show solutions in decimal form

θ = 2 πn, θ = π + 2 πn, θ = 0.84806 \dots + 2 πn, θ = π - 0.84806 \dots + 2 πn

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

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

csc (θ) = \frac{7}{3}

6 sin (θ) = 3

2 = sec (x)

What is the general solution for 2cos(2θ)-2=-3sin(θ) ?
The general solution for 2cos(2θ)-2=-3sin(θ) is θ=2pin,θ=pi+2pin,θ=0.84806…+2pin,θ=pi-0.84806…+2pin