English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

0<2cos(2x)-sqrt(3)

Solution

0 < 2 cos (2 x) - 3

Solution

- \frac{π}{12} + πn < x < \frac{π}{12} + πn

Interval Notation

(- \frac{π}{12} + πn, \frac{π}{12} + πn)

Decimal Notation

- 0.26179 \dots + πn < x < 0.26179 \dots + πn

Solution steps

0 < 2 cos (2 x) - 3

Switch sides

2 cos (2 x) - 3 > 0

Move

3

to the right side

2 cos (2 x) - 3 > 0

Add

3

to both sides

2 cos (2 x) - 3 + 3 > 0 + 3

Simplify

2 cos (2 x) > 3

2 cos (2 x) > 3

Divide both sides by

2

2 cos (2 x) > 3

Divide both sides by

2

\frac{2 cos ( 2 x )}{2} > \frac{3}{2}

Simplify

cos (2 x) > \frac{3}{2}

cos (2 x) > \frac{3}{2}

For

cos (x) > a

, if

- 1 \leq a < 1

then

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

- arccos (\frac{3}{2}) + 2 πn < 2 x < arccos (\frac{3}{2}) + 2 πn

a < u < b

then

a < u and u < b

- arccos (\frac{3}{2}) + 2 πn < 2 x and 2 x < arccos (\frac{3}{2}) + 2 πn

- arccos (\frac{3}{2}) + 2 πn < 2 x : x > - \frac{π}{12} + πn

- arccos (\frac{3}{2}) + 2 πn < 2 x

Switch sides

2 x > - arccos (\frac{3}{2}) + 2 πn

Simplify

- arccos (\frac{3}{2}) + 2 πn : - \frac{π}{6} + 2 πn

- arccos (\frac{3}{2}) + 2 πn

Use the following trivial identity:

arccos (\frac{3}{2}) = \frac{π}{6}

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}{6} + 2 πn

2 x > - \frac{π}{6} + 2 πn

Divide both sides by

2

2 x > - \frac{π}{6} + 2 πn

Divide both sides by

2

\frac{2 x}{2} > - \frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} > - \frac{\frac{π}{6}}{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}{6}}{2} + \frac{2 πn}{2} : - \frac{π}{12} + πn

- \frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{6}}{2} = \frac{π}{12}

\frac{\frac{π}{6}}{2}

Apply the fraction rule:

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

= \frac{π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= - \frac{π}{12} + πn

x > - \frac{π}{12} + πn

x > - \frac{π}{12} + πn

x > - \frac{π}{12} + πn

2 x < arccos (\frac{3}{2}) + 2 πn : x < \frac{π}{12} + πn

2 x < arccos (\frac{3}{2}) + 2 πn

Simplify

arccos (\frac{3}{2}) + 2 πn : \frac{π}{6} + 2 πn

arccos (\frac{3}{2}) + 2 πn

Use the following trivial identity:

arccos (\frac{3}{2}) = \frac{π}{6}

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}{6} + 2 πn

2 x < \frac{π}{6} + 2 πn

Divide both sides by

2

2 x < \frac{π}{6} + 2 πn

Divide both sides by

2

\frac{2 x}{2} < \frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} < \frac{\frac{π}{6}}{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}{6}}{2} + \frac{2 πn}{2} : \frac{π}{12} + πn

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

\frac{\frac{π}{6}}{2} = \frac{π}{12}

\frac{\frac{π}{6}}{2}

Apply the fraction rule:

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

= \frac{π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{12} + πn

x < \frac{π}{12} + πn

x < \frac{π}{12} + πn

x < \frac{π}{12} + πn

Combine the intervals

x > - \frac{π}{12} + πn and x < \frac{π}{12} + πn

Merge Overlapping Intervals

- \frac{π}{12} + πn < x < \frac{π}{12} + πn

Popular Examples

tan(x)tan(2x)>1

tan (x) tan (2 x) > 1

tan(a)<0

tan (a) < 0

1-sin(x)>= 0

1 - sin (x) \geq 0

5/4 sin(3/4 x-6)+1/4 <= 3/2

\frac{5}{4} sin (\frac{3}{4} x - 6) + \frac{1}{4} \leq \frac{3}{2}

-1/2 (10sin(2pi*60x))>-0.52

- \frac{1}{2} (10 sin (2 π \cdot 60 x)) > - 0.52