What is the general solution for 8cos(2θ)=-8cos(θ) ?

The general solution for 8cos(2θ)=-8cos(θ) is θ= pi/3+2pin,θ=(5pi)/3+2pin,θ=pi+2pin

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8cos(2θ)=-8cos(θ)

8 cos (2 θ) = - 8 cos (θ)

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

Solution steps

8 cos (2 θ) = - 8 cos (θ)

Subtract

- 8 cos (θ)

from both sides

8 cos (2 θ) + 8 cos (θ) = 0

Rewrite using trig identities

(- 1 + 2 cos^{2} (θ)) \cdot 8 + 8 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

arcsin (x) = \frac{π}{4}

arcsin (x) = \frac{π}{3}

sin^{2} (x) = 4

6 - sin (θ) = cos (2 θ)

sin (x) = - \frac{1}{2}, - π \leq x \leq π

What is the general solution for 8cos(2θ)=-8cos(θ) ?
The general solution for 8cos(2θ)=-8cos(θ) is θ= pi/3+2pin,θ=(5pi)/3+2pin,θ=pi+2pin