What is the general solution for 4cos(2θ)=0 ?

The general solution for 4cos(2θ)=0 is θ= pi/4+pin,θ=(3pi)/4+pin

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4cos(2θ)=0

4 cos (2 θ) = 0

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

Solution steps

4 cos (2 θ) = 0

Divide both sides by

4

cos (2 θ) = 0

General solutions for

cos (2 θ) = 0

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

Solve

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

Solve

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

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

\frac{1}{2} = sin (θ)

sin (2 x) - 2 sin (x) = 0

3 arcsin (x) = π

6 sec (θ) + 8 = - 4

cos^{2} (θ) - cos (θ) - 2 = 0

What is the general solution for 4cos(2θ)=0 ?
The general solution for 4cos(2θ)=0 is θ= pi/4+pin,θ=(3pi)/4+pin