What is the general solution for 4=-8cos(θ) ?

The general solution for 4=-8cos(θ) is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin

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4=-8cos(θ)

4 = - 8 cos (θ)

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

Solution steps

4 = - 8 cos (θ)

Switch sides

- 8 cos (θ) = 4

Divide both sides by

- 8

cos (θ) = - \frac{1}{2}

General solutions for

cos (θ) = - \frac{1}{2}

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

2 cos^{2} (θ) - 5 cos (θ) = 3

12 sin^{2} (x) - 20 sin (x) = - 8

6 arccos (4 x) = π

- 4 sin (x) cos (x) = 2 sin (x)

7 sin^{2} (θ) + 16 sin (θ) + 9 = 0

What is the general solution for 4=-8cos(θ) ?
The general solution for 4=-8cos(θ) is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin