English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sqrt(3)+2cos(2x)>= 0

Solution

3 + 2 cos (2 x) \geq 0

Solution

- \frac{5 π}{12} + πn \leq x \leq \frac{5 π}{12} + πn

Interval Notation

[- \frac{5 π}{12} + πn, \frac{5 π}{12} + πn]

Decimal Notation

- 1.30899 \dots + πn \leq x \leq 1.30899 \dots + πn

Solution steps

3 + 2 cos (2 x) \geq 0

Move

3

to the right side

3 + 2 cos (2 x) \geq 0

Subtract

3

from both sides

3 + 2 cos (2 x) - 3 \geq 0 - 3

Simplify

2 cos (2 x) \geq - 3

2 cos (2 x) \geq - 3

Divide both sides by

2

2 cos (2 x) \geq - 3

Divide both sides by

2

\frac{2 cos ( 2 x )}{2} \geq \frac{- 3}{2}

Simplify

cos (2 x) \geq - \frac{3}{2}

cos (2 x) \geq - \frac{3}{2}

For

cos (x) \geq a

, if

- 1 < a < 1

then

- arccos (a) + 2 πn \leq x \leq arccos (a) + 2 πn

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

a \leq u \leq b

then

a \leq u and u \leq b

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

- arccos (- \frac{3}{2}) + 2 πn \leq 2 x : x \geq - \frac{5 π}{12} + πn

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

Switch sides

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

Simplify

- arccos (- \frac{3}{2}) + 2 πn : - \frac{5 π}{6} + 2 πn

- arccos (- \frac{3}{2}) + 2 πn

Use the following trivial identity:

arccos (- \frac{3}{2}) = \frac{5 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{5 π}{6} + 2 πn

2 x \geq - \frac{5 π}{6} + 2 πn

Divide both sides by

2

2 x \geq - \frac{5 π}{6} + 2 πn

Divide both sides by

2

\frac{2 x}{2} \geq - \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} \geq - \frac{\frac{5 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{5 π}{6}}{2} + \frac{2 πn}{2} : - \frac{5 π}{12} + πn

- \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{6}}{2} = \frac{5 π}{12}

\frac{\frac{5 π}{6}}{2}

Apply the fraction rule:

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

= \frac{5 π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{5 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= - \frac{5 π}{12} + πn

x \geq - \frac{5 π}{12} + πn

x \geq - \frac{5 π}{12} + πn

x \geq - \frac{5 π}{12} + πn

2 x \leq arccos (- \frac{3}{2}) + 2 πn : x \leq \frac{5 π}{12} + πn

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

Simplify

arccos (- \frac{3}{2}) + 2 πn : \frac{5 π}{6} + 2 πn

arccos (- \frac{3}{2}) + 2 πn

Use the following trivial identity:

arccos (- \frac{3}{2}) = \frac{5 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{5 π}{6} + 2 πn

2 x \leq \frac{5 π}{6} + 2 πn

Divide both sides by

2

2 x \leq \frac{5 π}{6} + 2 πn

Divide both sides by

2

\frac{2 x}{2} \leq \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} \leq \frac{\frac{5 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{5 π}{6}}{2} + \frac{2 πn}{2} : \frac{5 π}{12} + πn

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

\frac{\frac{5 π}{6}}{2} = \frac{5 π}{12}

\frac{\frac{5 π}{6}}{2}

Apply the fraction rule:

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

= \frac{5 π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{5 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{5 π}{12} + πn

x \leq \frac{5 π}{12} + πn

x \leq \frac{5 π}{12} + πn

x \leq \frac{5 π}{12} + πn

Combine the intervals

x \geq - \frac{5 π}{12} + πn and x \leq \frac{5 π}{12} + πn

Merge Overlapping Intervals

- \frac{5 π}{12} + πn \leq x \leq \frac{5 π}{12} + πn

Popular Examples

4(cos(x))^2>1

4 (cos (x))^{2} > 1

sec(-5)sin(θ)>0

sec (- 5) sin (θ) > 0

2sin(x)+sqrt(3)>0

2 sin (x) + 3 > 0

cos^2(x)>= 1

cos^{2} (x) \geq 1

csc(θ)((sqrt(26))/5)tan(θ)<0

csc (θ) (\frac{26}{5}) tan (θ) < 0