What is the general solution for 2cos(4θ)-1=0 ?

The general solution for 2cos(4θ)-1=0 is θ= pi/(12)+(pin)/2 ,θ=(5pi}{12}+\frac{pin)/2

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2cos(4θ)-1=0

2 cos (4 θ) - 1 = 0

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

Solution steps

2 cos (4 θ) - 1 = 0

Move

1

to the right side

2 cos (4 θ) = 1

Divide both sides by

2

cos (4 θ) = \frac{1}{2}

General solutions for

cos (4 θ) = \frac{1}{2}

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

Solve

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

Solve

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

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

3 cos (x) = 8 tan (x)

1 + 4 sin^{2} (x) - 5 sin (x) + cos (2 x) = 0

cos (x) = \frac{9}{10}

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

2 cos (x) = tan (x)

What is the general solution for 2cos(4θ)-1=0 ?
The general solution for 2cos(4θ)-1=0 is θ= pi/(12)+(pin)/2 ,θ=(5pi}{12}+\frac{pin)/2