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Popular >

(9cos(t)-1/2)/4 <= 0

Solution

\frac{9 cos ( t ) - \frac{1}{2}}{4} \leq 0

Solution

arccos (\frac{1}{18}) + 2 πn \leq t \leq 2 π - arccos (\frac{1}{18}) + 2 πn

+2

Interval Notation

[arccos (\frac{1}{18}) + 2 πn, 2 π - arccos (\frac{1}{18}) + 2 πn]

Decimal Notation

1.51521 \dots + 2 πn \leq t \leq 4.76797 \dots + 2 πn

Solution steps

\frac{9 cos ( t ) - \frac{1}{2}}{4} \leq 0

Multiply both sides by

4

\frac{9 cos ( t ) - \frac{1}{2}}{4} \leq 0

Multiply both sides by

4

\frac{4 ( 9 cos ( t ) - \frac{1}{2} )}{4} \leq 0 \cdot 4

Simplify

9 cos (t) - \frac{1}{2} \leq 0

9 cos (t) - \frac{1}{2} \leq 0

Move

\frac{1}{2}

to the right side

9 cos (t) - \frac{1}{2} \leq 0

Add

\frac{1}{2}

to both sides

9 cos (t) - \frac{1}{2} + \frac{1}{2} \leq 0 + \frac{1}{2}

Simplify

9 cos (t) \leq \frac{1}{2}

9 cos (t) \leq \frac{1}{2}

Divide both sides by

9

9 cos (t) \leq \frac{1}{2}

Divide both sides by

9

\frac{9 cos ( t )}{9} \leq \frac{\frac{1}{2}}{9}

Simplify

\frac{9 cos ( t )}{9} \leq \frac{\frac{1}{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{1}{2}}{9} : \frac{1}{18}

\frac{\frac{1}{2}}{9}

Apply the fraction rule:

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

= \frac{1}{2 \cdot 9}

Multiply the numbers:

2 \cdot 9 = 18

= \frac{1}{18}

cos (t) \leq \frac{1}{18}

cos (t) \leq \frac{1}{18}

cos (t) \leq \frac{1}{18}

For

cos (x) \leq a

, if

- 1 < a < 1

then

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

arccos (\frac{1}{18}) + 2 πn \leq t \leq 2 π - arccos (\frac{1}{18}) + 2 πn

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