What is the general solution for 6cos(2θ)+3=0 ?

The general solution for 6cos(2θ)+3=0 is θ= pi/3+pin,θ=(2pi)/3+pin

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6cos(2θ)+3=0

6 cos (2 θ) + 3 = 0

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

Solution steps

6 cos (2 θ) + 3 = 0

Move

3

to the right side

6 cos (2 θ) = - 3

Divide both sides by

6

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

(sin (θ) + cos^{2} (θ))^{2} = 2 sin (θ)

\frac{2}{3} = sin (2 θ)

sin (3 x) - 2 = - 3 sin (3 x)

sin (a) = \frac{1}{5}

θ, cos (θ) = \frac{2}{3}

What is the general solution for 6cos(2θ)+3=0 ?
The general solution for 6cos(2θ)+3=0 is θ= pi/3+pin,θ=(2pi)/3+pin