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Popular >

(9cos(t)-1/2)/4 >= 4

Solution

\frac{9 cos ( t ) - \frac{1}{2}}{4} \geq 4

Solution

False for all t \in R

Solution steps

\frac{9 cos ( t ) - \frac{1}{2}}{4} \geq 4

Multiply both sides by

4

\frac{9 cos ( t ) - \frac{1}{2}}{4} \geq 4

Multiply both sides by

4

\frac{4 ( 9 cos ( t ) - \frac{1}{2} )}{4} \geq 4 \cdot 4

Simplify

9 cos (t) - \frac{1}{2} \geq 16

9 cos (t) - \frac{1}{2} \geq 16

Move

\frac{1}{2}

to the right side

9 cos (t) - \frac{1}{2} \geq 16

Add

\frac{1}{2}

to both sides

9 cos (t) - \frac{1}{2} + \frac{1}{2} \geq 16 + \frac{1}{2}

Simplify

9 cos (t) - \frac{1}{2} + \frac{1}{2} \geq 16 + \frac{1}{2}

Simplify

9 cos (t) - \frac{1}{2} + \frac{1}{2} : 9 cos (t)

9 cos (t) - \frac{1}{2} + \frac{1}{2}

Add similar elements:

- \frac{1}{2} + \frac{1}{2} \geq 0

= 9 cos (t)

Simplify

16 + \frac{1}{2} : \frac{33}{2}

16 + \frac{1}{2}

Convert element to fraction:

16 = \frac{16 \cdot 2}{2}

= \frac{16 \cdot 2}{2} + \frac{1}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{16 \cdot 2 + 1}{2}

16 \cdot 2 + 1 = 33

16 \cdot 2 + 1

Multiply the numbers:

16 \cdot 2 = 32

= 32 + 1

Add the numbers:

32 + 1 = 33

= 33

= \frac{33}{2}

9 cos (t) \geq \frac{33}{2}

9 cos (t) \geq \frac{33}{2}

9 cos (t) \geq \frac{33}{2}

Divide both sides by

9

9 cos (t) \geq \frac{33}{2}

Divide both sides by

9

\frac{9 cos ( t )}{9} \geq \frac{\frac{33}{2}}{9}

Simplify

\frac{9 cos ( t )}{9} \geq \frac{\frac{33}{2}}{9}

Simplify

\frac{9 cos ( t )}{9} : cos (t)

\frac{9 cos ( t )}{9}

Divide the numbers:

\frac{9}{9} = 1

= cos (t)

Simplify

\frac{\frac{33}{2}}{9} : \frac{11}{6}

\frac{\frac{33}{2}}{9}

Apply the fraction rule:

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

= \frac{33}{2 \cdot 9}

Multiply the numbers:

2 \cdot 9 = 18

= \frac{33}{18}

Cancel the common factor:

3

= \frac{11}{6}

cos (t) \geq \frac{11}{6}

cos (t) \geq \frac{11}{6}

cos (t) \geq \frac{11}{6}

Range of

cos (t) : - 1 \leq cos (t) \leq 1

Function range definition

The range of the basic

cos

function is

- 1 \leq cos (t) \leq 1

- 1 \leq cos (t) \leq 1

cos (t) \geq \frac{11}{6} and - 1 \leq cos (t) \leq 1 :

False

Let

y = cos (t)

Combine the intervals

y \geq \frac{11}{6} and - 1 \leq y \leq 1

Merge Overlapping Intervals

y \geq \frac{11}{6} and - 1 \leq y \leq 1

The intersection of two intervals is the set of numbers which are in both intervals

y \geq \frac{11}{6}

and

- 1 \leq y \leq 1

False for all y \in R

False for all y \in R

No Solution for t \in R

False for all t \in R

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