English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3cos(2x)+2< 5/2

Solution

3 cos (2 x) + 2 < \frac{5}{2}

Solution

\frac{arccos ( \frac{1}{6} )}{2} + πn < x < \frac{2 π - arccos ( \frac{1}{6} )}{2} + πn

Interval Notation

(\frac{arccos ( \frac{1}{6} )}{2} + πn, \frac{2 π - arccos ( \frac{1}{6} )}{2} + πn)

Decimal Notation

0.70167 \dots + πn < x < 2.43991 \dots + πn

Solution steps

3 cos (2 x) + 2 < \frac{5}{2}

Move

2

to the right side

3 cos (2 x) + 2 < \frac{5}{2}

Subtract

2

from both sides

3 cos (2 x) + 2 - 2 < \frac{5}{2} - 2

Simplify

3 cos (2 x) + 2 - 2 < \frac{5}{2} - 2

Simplify

3 cos (2 x) + 2 - 2 : 3 cos (2 x)

3 cos (2 x) + 2 - 2

Add similar elements:

2 - 2 < 0

= 3 cos (2 x)

Simplify

\frac{5}{2} - 2 : \frac{1}{2}

\frac{5}{2} - 2

Convert element to fraction:

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

= - \frac{2 \cdot 2}{2} + \frac{5}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{- 2 \cdot 2 + 5}{2}

- 2 \cdot 2 + 5 = 1

- 2 \cdot 2 + 5

Multiply the numbers:

2 \cdot 2 = 4

= - 4 + 5

Add/Subtract the numbers:

- 4 + 5 = 1

= 1

= \frac{1}{2}

3 cos (2 x) < \frac{1}{2}

3 cos (2 x) < \frac{1}{2}

3 cos (2 x) < \frac{1}{2}

Divide both sides by

3

3 cos (2 x) < \frac{1}{2}

Divide both sides by

3

\frac{3 cos ( 2 x )}{3} < \frac{\frac{1}{2}}{3}

Simplify

\frac{3 cos ( 2 x )}{3} < \frac{\frac{1}{2}}{3}

Simplify

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

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

Divide the numbers:

\frac{3}{3} = 1

= cos (2 x)

Simplify

\frac{\frac{1}{2}}{3} : \frac{1}{6}

\frac{\frac{1}{2}}{3}

Apply the fraction rule:

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

= \frac{1}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{1}{6}

cos (2 x) < \frac{1}{6}

cos (2 x) < \frac{1}{6}

cos (2 x) < \frac{1}{6}

For

cos (x) < a

, if

- 1 < a \leq 1

then

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

arccos (\frac{1}{6}) + 2 πn < 2 x < 2 π - arccos (\frac{1}{6}) + 2 πn

a < u < b

then

a < u and u < b

arccos (\frac{1}{6}) + 2 πn < 2 x and 2 x < 2 π - arccos (\frac{1}{6}) + 2 πn

arccos (\frac{1}{6}) + 2 πn < 2 x : x > \frac{arccos ( \frac{1}{6} )}{2} + πn

arccos (\frac{1}{6}) + 2 πn < 2 x

Switch sides

2 x > arccos (\frac{1}{6}) + 2 πn

Divide both sides by

2

2 x > arccos (\frac{1}{6}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} > \frac{arccos ( \frac{1}{6} )}{2} + \frac{2 πn}{2}

Simplify

x > \frac{arccos ( \frac{1}{6} )}{2} + πn

x > \frac{arccos ( \frac{1}{6} )}{2} + πn

2 x < 2 π - arccos (\frac{1}{6}) + 2 πn : x < \frac{2 π - arccos ( \frac{1}{6} )}{2} + πn

2 x < 2 π - arccos (\frac{1}{6}) + 2 πn

Divide both sides by

2

2 x < 2 π - arccos (\frac{1}{6}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} < \frac{2 π}{2} - \frac{arccos ( \frac{1}{6} )}{2} + \frac{2 πn}{2}

Simplify

x < π - \frac{arccos ( \frac{1}{6} )}{2} + πn

Simplify

π - \frac{arccos ( \frac{1}{6} )}{2} : \frac{2 π - arccos ( \frac{1}{6} )}{2}

π - \frac{arccos ( \frac{1}{6} )}{2}

Convert element to fraction:

π = \frac{π 2}{2}

= \frac{π 2}{2} - \frac{arccos ( \frac{1}{6} )}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{π 2 - arccos ( \frac{1}{6} )}{2}

x < \frac{2 π - arccos ( \frac{1}{6} )}{2} + πn

x < \frac{2 π - arccos ( \frac{1}{6} )}{2} + πn

Combine the intervals

x > \frac{arccos ( \frac{1}{6} )}{2} + πn and x < \frac{2 π - arccos ( \frac{1}{6} )}{2} + πn

Merge Overlapping Intervals

\frac{arccos ( \frac{1}{6} )}{2} + πn < x < \frac{2 π - arccos ( \frac{1}{6} )}{2} + πn

Popular Examples

cos(x+1)<= 5

cos (x + 1) \leq 5

cos(x)<= pi

cos (x) \leq π

cos(x)-sin(x)>= 1

cos (x) - sin (x) \geq 1

sin(x)-cos(x)<0

sin (x) - cos (x) < 0

cos(3x)>= 3

cos (3 x) \geq 3