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

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

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2*cos(4x)-1=0

2 \cdot cos (4 x) - 1 = 0

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

Solution steps

2 cos (4 x) - 1 = 0

Move

1

to the right side

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}

2 cos^{2} (θ) - 3 cos (θ) = 2

arctan (0.2 x) = 4 0^{\circ}

sin (x) - sin (2 x) + sin (3 x) = 0

sin (x) = \frac{- 2 5}{5}

tan (2 x - 1) = 0

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