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Popular >

arccos((x-5)/(10))-pi/3 >= 0

Solution

arccos (\frac{x - 5}{10}) - \frac{π}{3} \geq 0

Solution

- 5 \leq x \leq 10

Interval Notation

[- 5, 10]

Solution steps

arccos (\frac{x - 5}{10}) - \frac{π}{3} \geq 0

Move

\frac{π}{3}

to the right side

arccos (\frac{x - 5}{10}) - \frac{π}{3} \geq 0

Add

\frac{π}{3}

to both sides

arccos (\frac{x - 5}{10}) - \frac{π}{3} + \frac{π}{3} \geq 0 + \frac{π}{3}

Simplify

arccos (\frac{x - 5}{10}) \geq \frac{π}{3}

arccos (\frac{x - 5}{10}) \geq \frac{π}{3}

arccos (x) \geq a

then

x \leq cos (a)

\frac{x - 5}{10} \leq cos (\frac{π}{3})

cos (\frac{π}{3}) = \frac{1}{2}

cos (\frac{π}{3})

Use the following trivial identity:

cos (\frac{π}{3}) = \frac{1}{2}

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{1}{2}

\frac{x - 5}{10} \leq \frac{1}{2}

\frac{x - 5}{10} \leq \frac{1}{2} : x \leq 10

\frac{x - 5}{10} \leq \frac{1}{2}

Multiply both sides by

10

\frac{x - 5}{10} \leq \frac{1}{2}

Multiply both sides by

10

\frac{x - 5}{10} \cdot 10 \leq \frac{1}{2} \cdot 10

Simplify

\frac{x - 5}{10} \cdot 10 \leq \frac{1}{2} \cdot 10

Simplify

\frac{x - 5}{10} \cdot 10 : x - 5

\frac{x - 5}{10} \cdot 10

Multiply fractions:

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

= \frac{( x - 5 ) \cdot 10}{10}

Cancel the common factor:

10

= x - 5

Simplify

\frac{1}{2} \cdot 10 : 5

\frac{1}{2} \cdot 10

Convert element to fraction:

10 = \frac{10}{1}

= \frac{1}{2} \cdot \frac{10}{1}

Cross-cancel common factor:

2

= \frac{5}{1}

Apply the fraction rule:

\frac{a}{1} = a

= 5

x - 5 \leq 5

x - 5 \leq 5

x - 5 \leq 5

Move

5

to the right side

x - 5 \leq 5

Add

5

to both sides

x - 5 + 5 \leq 5 + 5

Simplify

x \leq 10

x \leq 10

x \leq 10

Domain of

arccos (\frac{x - 5}{10}) : - 5 \leq x \leq 15

Domain definition

Find known functions domain restrictions:

- 5 \leq x \leq 15

arccos (f (x)) \Rightarrow - 1 \leq f (x) \leq 1

Solve

- 1 \leq \frac{x - 5}{10} \leq 1 : - 5 \leq x \leq 15

- 1 \leq \frac{x - 5}{10} \leq 1

a \leq u \leq b

then

a \leq u and u \leq b

- 1 \leq \frac{x - 5}{10} and \frac{x - 5}{10} \leq 1

- 1 \leq \frac{x - 5}{10} : x \geq - 5

- 1 \leq \frac{x - 5}{10}

Switch sides

\frac{x - 5}{10} \geq - 1

Multiply both sides by

10

\frac{x - 5}{10} \geq - 1

Multiply both sides by

10

\frac{10 ( x - 5 )}{10} \geq 10 (- 1)

Simplify

x - 5 \geq - 10

x - 5 \geq - 10

Move

5

to the right side

x - 5 \geq - 10

Add

5

to both sides

x - 5 + 5 \geq - 10 + 5

Simplify

x \geq - 5

x \geq - 5

\frac{x - 5}{10} \leq 1 : x \leq 15

\frac{x - 5}{10} \leq 1

Multiply both sides by

10

\frac{x - 5}{10} \leq 1

Multiply both sides by

10

\frac{10 ( x - 5 )}{10} \leq 1 \cdot 10

Simplify

x - 5 \leq 10

x - 5 \leq 10

Move

5

to the right side

x - 5 \leq 10

Add

5

to both sides

x - 5 + 5 \leq 10 + 5

Simplify

x \leq 15

x \leq 15

Combine the intervals

x \geq - 5 and x \leq 15

Merge Overlapping Intervals

x \geq - 5 and x \leq 15

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

x \geq - 5

and

x \leq 15

- 5 \leq x \leq 15

- 5 \leq x \leq 15

The function domain

- 5 \leq x \leq 15

Combine the intervals

x \leq 10 and - 5 \leq x \leq 15

Merge Overlapping Intervals

x \leq 10 and - 5 \leq x \leq 15

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

x \leq 10

and

- 5 \leq x \leq 15

- 5 \leq x \leq 10

- 5 \leq x \leq 10

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