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

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

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3cos(θ)+cos(2θ)=-2

3 cos (θ) + cos (2 θ) = - 2

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

Solution steps

3 cos (θ) + cos (2 θ) = - 2

Subtract

- 2

from both sides

3 cos (θ) + cos (2 θ) + 2 = 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

15 sin (x) + 7 = sin (x)

3 sin^{2} (x) + 2 sin (x) - 1 = 0

2 cos (θ) = 3

- 73 tan (θ) + 2 = 9

tan^{2} (x) + 2 sec (x) = - 1

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