What is the general solution for 3+8cos(2θ)=-1 ?

The general solution for 3+8cos(2θ)=-1 is θ= pi/3+pin,θ=(2pi)/3+pin

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3+8cos(2θ)=-1

3 + 8 cos (2 θ) = - 1

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

Solution steps

3 + 8 cos (2 θ) = - 1

Move

3

to the right side

8 cos (2 θ) = - 4

Divide both sides by

8

cos (2 θ) = - \frac{1}{2}

General solutions for

cos (2 θ) = - \frac{1}{2}

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

Solve

2 θ = \frac{2 π}{3} + 2 πn : θ = \frac{π}{3} + πn

Solve

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

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

- 4 cos^{2} (x) + 6 cos (x) = 2

sec (θ) = - \frac{4}{3}

tan (2 x) = tan (201 8^{\circ})

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

\frac{6}{π} - 4 sin (2 x) = 0

What is the general solution for 3+8cos(2θ)=-1 ?
The general solution for 3+8cos(2θ)=-1 is θ= pi/3+pin,θ=(2pi)/3+pin