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Popular >

3cos(x)+11/2 >4

Solution

3 cos (x) + \frac{11}{2} > 4

Solution

- \frac{2 π}{3} + 2 πn < x < \frac{2 π}{3} + 2 πn

Interval Notation

(- \frac{2 π}{3} + 2 πn, \frac{2 π}{3} + 2 πn)

Decimal Notation

- 2.09439 \dots + 2 πn < x < 2.09439 \dots + 2 πn

Solution steps

3 cos (x) + \frac{11}{2} > 4

Move

\frac{11}{2}

to the right side

3 cos (x) + \frac{11}{2} > 4

Subtract

\frac{11}{2}

from both sides

3 cos (x) + \frac{11}{2} - \frac{11}{2} > 4 - \frac{11}{2}

Simplify

3 cos (x) + \frac{11}{2} - \frac{11}{2} > 4 - \frac{11}{2}

Simplify

3 cos (x) + \frac{11}{2} - \frac{11}{2} : 3 cos (x)

3 cos (x) + \frac{11}{2} - \frac{11}{2}

Add similar elements:

\frac{11}{2} - \frac{11}{2} > 0

= 3 cos (x)

Simplify

4 - \frac{11}{2} : - \frac{3}{2}

4 - \frac{11}{2}

Convert element to fraction:

4 = \frac{4 \cdot 2}{2}

= \frac{4 \cdot 2}{2} - \frac{11}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{4 \cdot 2 - 11}{2}

4 \cdot 2 - 11 = - 3

4 \cdot 2 - 11

Multiply the numbers:

4 \cdot 2 = 8

= 8 - 11

Subtract the numbers:

8 - 11 = - 3

= - 3

= \frac{- 3}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{3}{2}

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

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

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

Divide both sides by

3

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

Divide both sides by

3

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

Simplify

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

Simplify

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

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

Divide the numbers:

\frac{3}{3} = 1

= cos (x)

Simplify

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

\frac{- \frac{3}{2}}{3}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{\frac{3}{2}}{3}

Apply the fraction rule:

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

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

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

Multiply the numbers:

2 \cdot 3 = 6

= - \frac{3}{6}

Cancel the common factor:

3

= - \frac{1}{2}

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

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

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

For

cos (x) > a

, if

- 1 \leq a < 1

then

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

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

Simplify

- arccos (- \frac{1}{2}) : - \frac{2 π}{3}

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

Use the following trivial identity:

arccos (- \frac{1}{2}) = \frac{2 π}{3}

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{2 π}{3}

Simplify

arccos (- \frac{1}{2}) : \frac{2 π}{3}

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

Use the following trivial identity:

arccos (- \frac{1}{2}) = \frac{2 π}{3}

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{2 π}{3}

- \frac{2 π}{3} + 2 πn < x < \frac{2 π}{3} + 2 πn

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