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

Q: What is the general solution for 2cos(2θ)-1=0,0<= θ<= 2pi ?

The general solution for 2cos(2θ)-1=0,0<= θ<= 2pi is θ= pi/6 ,θ=(5pi)/6 ,θ=(7pi)/6 ,θ=(11pi)/6

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2cos(2θ)-1=0,0<= θ<= 2pi

Solution

2 cos (2 θ) - 1 = 0, 0 \leq θ \leq 2 π

Solution

θ = \frac{π}{6}, θ = \frac{5 π}{6}, θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

Degrees

θ = 3 0^{\circ}, θ = 15 0^{\circ}, θ = 21 0^{\circ}, θ = 33 0^{\circ}

Solution steps

2 cos (2 θ) - 1 = 0, 0 \leq θ \leq 2 π

Move

1

to the right side

2 cos (2 θ) - 1 = 0

Add

1

to both sides

2 cos (2 θ) - 1 + 1 = 0 + 1

Simplify

2 cos (2 θ) = 1

2 cos (2 θ) = 1

Divide both sides by

2

2 cos (2 θ) = 1

Divide both sides by

2

\frac{2 cos ( 2 θ )}{2} = \frac{1}{2}

Simplify

cos (2 θ) = \frac{1}{2}

cos (2 θ) = \frac{1}{2}

General solutions for

cos (2 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}

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

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

Solve

2 θ = \frac{π}{3} + 2 πn : θ = \frac{π}{6} + πn

2 θ = \frac{π}{3} + 2 πn

Divide both sides by

2

2 θ = \frac{π}{3} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{\frac{π}{3}}{2} + \frac{2 πn}{2} : \frac{π}{6} + πn

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

\frac{\frac{π}{3}}{2} = \frac{π}{6}

\frac{\frac{π}{3}}{2}

Apply the fraction rule:

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

= \frac{π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{π}{6}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{6} + πn

θ = \frac{π}{6} + πn

θ = \frac{π}{6} + πn

θ = \frac{π}{6} + πn

Solve

2 θ = \frac{5 π}{3} + 2 πn : θ = \frac{5 π}{6} + πn

2 θ = \frac{5 π}{3} + 2 πn

Divide both sides by

2

2 θ = \frac{5 π}{3} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{\frac{5 π}{3}}{2} + \frac{2 πn}{2} : \frac{5 π}{6} + πn

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

\frac{\frac{5 π}{3}}{2} = \frac{5 π}{6}

\frac{\frac{5 π}{3}}{2}

Apply the fraction rule:

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

= \frac{5 π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{5 π}{6}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{5 π}{6} + πn

θ = \frac{5 π}{6} + πn

θ = \frac{5 π}{6} + πn

θ = \frac{5 π}{6} + πn

θ = \frac{π}{6} + πn, θ = \frac{5 π}{6} + πn

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{π}{6}, θ = \frac{5 π}{6}, θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos(2θ)-1=0,0<= θ<= 2pi ?
The general solution for 2cos(2θ)-1=0,0<= θ<= 2pi is θ= pi/6 ,θ=(5pi)/6 ,θ=(7pi)/6 ,θ=(11pi)/6