What is the general solution for 1+cos(2θ)=-3sin(θ) ?

The general solution for 1+cos(2θ)=-3sin(θ) is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin

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1+cos(2θ)=-3sin(θ)

1 + cos (2 θ) = - 3 sin (θ)

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

Solution steps

1 + cos (2 θ) = - 3 sin (θ)

Subtract

- 3 sin (θ)

from both sides

1 + cos (2 θ) + 3 sin (θ) = 0

Rewrite using trig identities

2 - 2 sin^{2} (θ) + 3 sin (θ) = 0

Solve by substitution

sin (θ) = - \frac{1}{2}, sin (θ) = 2

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

sin (θ) = 2 :

No Solution

Combine all the solutions

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

2 cos (θ) + sin (2 θ) = 0

tan (x) = \frac{6 \cdot 3}{6}

cos^{2} (x) + sin^{2} (x) - 2 sin (x) = 0

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

2 + cos (2 x) = - 3 cos (x)

What is the general solution for 1+cos(2θ)=-3sin(θ) ?
The general solution for 1+cos(2θ)=-3sin(θ) is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin