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Popular >

cos(2x)> 1/(sqrt(2))

Solution

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

Solution

- \frac{π}{8} + πn < x < \frac{π}{8} + πn

Interval Notation

(- \frac{π}{8} + πn, \frac{π}{8} + πn)

Decimal Notation

- 0.39269 \dots + πn < x < 0.39269 \dots + πn

Solution steps

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

Simplify

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2}{2}

cos (2 x) > \frac{2}{2}

For

cos (x) > a

, if

- 1 \leq a < 1

then

- arccos (a) + 2 πn < x < arccos (a) + 2 πn

- arccos (\frac{2}{2}) + 2 πn < 2 x < arccos (\frac{2}{2}) + 2 πn

a < u < b

then

a < u and u < b

- arccos (\frac{2}{2}) + 2 πn < 2 x and 2 x < arccos (\frac{2}{2}) + 2 πn

- arccos (\frac{2}{2}) + 2 πn < 2 x : x > - \frac{π}{8} + πn

- arccos (\frac{2}{2}) + 2 πn < 2 x

Switch sides

2 x > - arccos (\frac{2}{2}) + 2 πn

Simplify

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

- arccos (\frac{2}{2}) + 2 πn

Use the following trivial identity:

arccos (\frac{2}{2}) = \frac{π}{4}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= - \frac{π}{4} + 2 πn

2 x > - \frac{π}{4} + 2 πn

Divide both sides by

2

2 x > - \frac{π}{4} + 2 πn

Divide both sides by

2

\frac{2 x}{2} > - \frac{\frac{π}{4}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} > - \frac{\frac{π}{4}}{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{\frac{π}{4}}{2} + \frac{2 πn}{2} : - \frac{π}{8} + πn

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

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

Apply the fraction rule:

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

= \frac{π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= - \frac{π}{8} + πn

x > - \frac{π}{8} + πn

x > - \frac{π}{8} + πn

x > - \frac{π}{8} + πn

2 x < arccos (\frac{2}{2}) + 2 πn : x < \frac{π}{8} + πn

2 x < arccos (\frac{2}{2}) + 2 πn

Simplify

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

arccos (\frac{2}{2}) + 2 πn

Use the following trivial identity:

arccos (\frac{2}{2}) = \frac{π}{4}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{π}{4} + 2 πn

2 x < \frac{π}{4} + 2 πn

Divide both sides by

2

2 x < \frac{π}{4} + 2 πn

Divide both sides by

2

\frac{2 x}{2} < \frac{\frac{π}{4}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} < \frac{\frac{π}{4}}{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{\frac{π}{4}}{2} + \frac{2 πn}{2} : \frac{π}{8} + πn

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

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

Apply the fraction rule:

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

= \frac{π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{8} + πn

x < \frac{π}{8} + πn

x < \frac{π}{8} + πn

x < \frac{π}{8} + πn

Combine the intervals

x > - \frac{π}{8} + πn and x < \frac{π}{8} + πn

Merge Overlapping Intervals

- \frac{π}{8} + πn < x < \frac{π}{8} + πn

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