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

The general solution for cos(2x)=cos(1/2) is x= 1/4+pin,x=pi-1/4+pin

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cos(2x)=cos(1/2)

Solution

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

Solution

x = \frac{1}{4} + πn, x = π - \frac{1}{4} + πn

Degrees

x = 14.32394 \dots^{\circ} + 18 0^{\circ} n, x = 165.67605 \dots^{\circ} + 18 0^{\circ} n

Solution steps

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

Apply trig inverse properties

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

General solutions for

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

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

2 x = arccos (cos (\frac{1}{2})) + 2 πn, 2 x = 2 π - arccos (cos (\frac{1}{2})) + 2 πn

2 x = arccos (cos (\frac{1}{2})) + 2 πn, 2 x = 2 π - arccos (cos (\frac{1}{2})) + 2 πn

Solve

2 x = arccos (cos (\frac{1}{2})) + 2 πn : x = \frac{1}{4} + πn

2 x = arccos (cos (\frac{1}{2})) + 2 πn

Divide both sides by

2

2 x = arccos (cos (\frac{1}{2})) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{arccos ( cos ( \frac{1}{2} ) )}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{arccos ( cos ( \frac{1}{2} ) )}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{arccos ( cos ( \frac{1}{2} ) )}{2} + \frac{2 πn}{2} : \frac{1}{4} + πn

\frac{arccos ( cos ( \frac{1}{2} ) )}{2} + \frac{2 πn}{2}

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

\frac{arccos ( cos ( \frac{1}{2} ) )}{2}

arccos (cos (\frac{1}{2})) = \frac{1}{2}

arccos (cos (\frac{1}{2}))

For

0 \leq x \leq π, a rccos (cos (x)) = x

0 \leq \frac{1}{2} \leq π

= \frac{1}{2}

= \frac{\frac{1}{2}}{2}

Apply the fraction rule:

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

= \frac{1}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{1}{4}

= \frac{1}{4} + \frac{2 πn}{2}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{1}{4} + πn

x = \frac{1}{4} + πn

x = \frac{1}{4} + πn

x = \frac{1}{4} + πn

Solve

2 x = 2 π - arccos (cos (\frac{1}{2})) + 2 πn : x = π - \frac{1}{4} + πn

2 x = 2 π - arccos (cos (\frac{1}{2})) + 2 πn

Divide both sides by

2

2 x = 2 π - arccos (cos (\frac{1}{2})) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{2 π}{2} - \frac{arccos ( cos ( \frac{1}{2} ) )}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{2 π}{2} - \frac{arccos ( cos ( \frac{1}{2} ) )}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{2 π}{2} - \frac{arccos ( cos ( \frac{1}{2} ) )}{2} + \frac{2 πn}{2} : π - \frac{1}{4} + πn

\frac{2 π}{2} - \frac{arccos ( cos ( \frac{1}{2} ) )}{2} + \frac{2 πn}{2}

\frac{2 π}{2} = π

\frac{2 π}{2}

Divide the numbers:

\frac{2}{2} = 1

= π

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

\frac{arccos ( cos ( \frac{1}{2} ) )}{2}

arccos (cos (\frac{1}{2})) = \frac{1}{2}

arccos (cos (\frac{1}{2}))

For

0 \leq x \leq π, a rccos (cos (x)) = x

0 \leq \frac{1}{2} \leq π

= \frac{1}{2}

= \frac{\frac{1}{2}}{2}

Apply the fraction rule:

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

= \frac{1}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{1}{4}

= π - \frac{1}{4} + \frac{2 πn}{2}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= π - \frac{1}{4} + πn

x = π - \frac{1}{4} + πn

x = π - \frac{1}{4} + πn

x = π - \frac{1}{4} + πn

x = \frac{1}{4} + πn, x = π - \frac{1}{4} + πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for cos(2x)=cos(1/2) ?
The general solution for cos(2x)=cos(1/2) is x= 1/4+pin,x=pi-1/4+pin