What is the general solution for 2cos(4x)=1 ?

The general solution for 2cos(4x)=1 is x= pi/(12)+(pin)/2 ,x=(5pi}{12}+\frac{pin)/2

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2cos(4x)=1

2 cos (4 x) = 1

x = \frac{π}{12} + \frac{πn}{2}, x = \frac{5 π}{12} + \frac{πn}{2}

Solution steps

2 cos (4 x) = 1

Divide both sides by

2

cos (4 x) = \frac{1}{2}

General solutions for

cos (4 x) = \frac{1}{2}

4 x = \frac{π}{3} + 2 πn, 4 x = \frac{5 π}{3} + 2 πn

Solve

4 x = \frac{π}{3} + 2 πn : x = \frac{π}{12} + \frac{πn}{2}

Solve

4 x = \frac{5 π}{3} + 2 πn : x = \frac{5 π}{12} + \frac{πn}{2}

x = \frac{π}{12} + \frac{πn}{2}, x = \frac{5 π}{12} + \frac{πn}{2}

cos (2 x) = 2 sin (2 x)

4 cos^{2} (x) - 1 = 1

2 sin (θ) - 1 = 0

4 cos^{2} (x) + 5 cos (x) - 6 = 0

3 cos (2 θ) = - 3 cos (θ)

What is the general solution for 2cos(4x)=1 ?
The general solution for 2cos(4x)=1 is x= pi/(12)+(pin)/2 ,x=(5pi}{12}+\frac{pin)/2