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

The general solution for 2cos(4θ)=0 is θ= pi/8+(pin)/2 ,θ=(3pi)/8+(pin)/2

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2cos(4θ)=0

2 cos (4 θ) = 0

θ = \frac{π}{8} + \frac{πn}{2}, θ = \frac{3 π}{8} + \frac{πn}{2}

Solution steps

2 cos (4 θ) = 0

Divide both sides by

2

cos (4 θ) = 0

General solutions for

cos (4 θ) = 0

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

Solve

4 θ = \frac{π}{2} + 2 πn : θ = \frac{π}{8} + \frac{πn}{2}

Solve

4 θ = \frac{3 π}{2} + 2 πn : θ = \frac{3 π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}, θ = \frac{3 π}{8} + \frac{πn}{2}

\frac{1032}{3145} sin (4 x) + \frac{2196}{3145} cos (4 x) = 0

6 - 2 sin (θ) - 8 sin^{2} (θ) = 0

cos (2 θ) = cos (3 θ)

θ, 2 sin (θ) = 1.5

cos (θ) = - 2109

What is the general solution for 2cos(4θ)=0 ?
The general solution for 2cos(4θ)=0 is θ= pi/8+(pin)/2 ,θ=(3pi)/8+(pin)/2