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

Solution

2 cos (4 θ) - 1 = 0

Solution

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

Degrees

θ = 1 5^{\circ} + 9 0^{\circ} n, θ = 7 5^{\circ} + 9 0^{\circ} n

Solution steps

2 cos (4 θ) - 1 = 0

Move

1

to the right side

2 cos (4 θ) - 1 = 0

Add

1

to both sides

2 cos (4 θ) - 1 + 1 = 0 + 1

Simplify

2 cos (4 θ) = 1

2 cos (4 θ) = 1

Divide both sides by

2

2 cos (4 θ) = 1

Divide both sides by

2

\frac{2 cos ( 4 θ )}{2} = \frac{1}{2}

Simplify

cos (4 θ) = \frac{1}{2}

cos (4 θ) = \frac{1}{2}

General solutions for

cos (4 θ) = \frac{1}{2}

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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}

4 θ = \frac{π}{3} + 2 πn

Divide both sides by

4

4 θ = \frac{π}{3} + 2 πn

Divide both sides by

4

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

Simplify

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

Simplify

\frac{4 θ}{4} : θ

\frac{4 θ}{4}

Divide the numbers:

\frac{4}{4} = 1

= θ

Simplify

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

\frac{\frac{π}{3}}{4} + \frac{2 πn}{4}

\frac{\frac{π}{3}}{4} = \frac{π}{12}

\frac{\frac{π}{3}}{4}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{π}{3 \cdot 4}

Multiply the numbers:

3 \cdot 4 = 12

= \frac{π}{12}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

Cancel the common factor:

2

= \frac{πn}{2}

= \frac{π}{12} + \frac{πn}{2}

θ = \frac{π}{12} + \frac{πn}{2}

θ = \frac{π}{12} + \frac{πn}{2}

θ = \frac{π}{12} + \frac{πn}{2}

Solve

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

4 θ = \frac{5 π}{3} + 2 πn

Divide both sides by

4

4 θ = \frac{5 π}{3} + 2 πn

Divide both sides by

4

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

Simplify

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

Simplify

\frac{4 θ}{4} : θ

\frac{4 θ}{4}

Divide the numbers:

\frac{4}{4} = 1

= θ

Simplify

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

\frac{\frac{5 π}{3}}{4} + \frac{2 πn}{4}

\frac{\frac{5 π}{3}}{4} = \frac{5 π}{12}

\frac{\frac{5 π}{3}}{4}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{5 π}{3 \cdot 4}

Multiply the numbers:

3 \cdot 4 = 12

= \frac{5 π}{12}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

Cancel the common factor:

2

= \frac{πn}{2}

= \frac{5 π}{12} + \frac{πn}{2}

θ = \frac{5 π}{12} + \frac{πn}{2}

θ = \frac{5 π}{12} + \frac{πn}{2}

θ = \frac{5 π}{12} + \frac{πn}{2}

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

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Frequently Asked Questions (FAQ)

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