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

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

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3cos(2θ)=-3cos(θ)

3 cos (2 θ) = - 3 cos (θ)

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn, θ = π + 2 πn

Solution steps

3 cos (2 θ) = - 3 cos (θ)

Subtract

- 3 cos (θ)

from both sides

3 cos (2 θ) + 3 cos (θ) = 0

Rewrite using trig identities

(- 1 + 2 cos^{2} (θ)) \cdot 3 + 3 cos (θ) = 0

Solve by substitution

cos (θ) = \frac{1}{2}, cos (θ) = - 1

cos (θ) = \frac{1}{2} : θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

cos (θ) = - 1 : θ = π + 2 πn

Combine all the solutions

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn, θ = π + 2 πn

csc^{2} (θ) + 3 csc (θ) + 2 = 0

cos^{2} (x) = 3 - 5 sin (x)

cos^{2} (θ) - sin^{2} (θ) = - 1

8 sin (\frac{x}{2}) + 8 cos (x) = 0

tan (\frac{θ}{2} + \frac{π}{6}) = 1

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