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Popular Trigonometry >

1/2 sin(x)-5<=-1/2

Solution

\frac{1}{2} sin (x) - 5 \leq - \frac{1}{2}

Solution

True for all x \in R

Interval Notation

(- \infty, \infty)

Solution steps

\frac{1}{2} sin (x) - 5 \leq - \frac{1}{2}

Move

5

to the right side

\frac{1}{2} sin (x) - 5 \leq - \frac{1}{2}

Add

5

to both sides

\frac{1}{2} sin (x) - 5 + 5 \leq - \frac{1}{2} + 5

Simplify

\frac{1}{2} sin (x) - 5 + 5 \leq - \frac{1}{2} + 5

Simplify

\frac{1}{2} sin (x) - 5 + 5 : \frac{1}{2} sin (x)

\frac{1}{2} sin (x) - 5 + 5

Add similar elements:

- 5 + 5 \leq 0

= \frac{1}{2} sin (x)

Simplify

- \frac{1}{2} + 5 : \frac{9}{2}

- \frac{1}{2} + 5

Convert element to fraction:

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

= \frac{5 \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{5 \cdot 2 - 1}{2}

5 \cdot 2 - 1 = 9

5 \cdot 2 - 1

Multiply the numbers:

5 \cdot 2 = 10

= 10 - 1

Subtract the numbers:

10 - 1 = 9

= 9

= \frac{9}{2}

\frac{1}{2} sin (x) \leq \frac{9}{2}

\frac{1}{2} sin (x) \leq \frac{9}{2}

\frac{1}{2} sin (x) \leq \frac{9}{2}

Multiply both sides by

2

\frac{1}{2} sin (x) \leq \frac{9}{2}

Multiply both sides by

2

2 \cdot \frac{1}{2} sin (x) \leq \frac{9 \cdot 2}{2}

Simplify

2 \cdot \frac{1}{2} sin (x) \leq \frac{9 \cdot 2}{2}

Simplify

2 \cdot \frac{1}{2} sin (x) : sin (x)

2 \cdot \frac{1}{2} sin (x)

Multiply fractions:

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

= \frac{1 \cdot 2}{2} sin (x)

Cancel the common factor:

2

= sin (x) \cdot 1

Multiply:

sin (x) \cdot 1 = sin (x)

= sin (x)

Simplify

\frac{9 \cdot 2}{2} : 9

\frac{9 \cdot 2}{2}

Multiply the numbers:

9 \cdot 2 = 18

= \frac{18}{2}

Divide the numbers:

\frac{18}{2} = 9

= 9

sin (x) \leq 9

sin (x) \leq 9

sin (x) \leq 9

Range of

sin (x) : - 1 \leq sin (x) \leq 1

Function range definition

The range of the basic

sin

function is

- 1 \leq sin (x) \leq 1

- 1 \leq sin (x) \leq 1

sin (x) \leq 9 and - 1 \leq sin (x) \leq 1 : - 1 \leq sin (x) \leq 1

Let

y = sin (x)

Combine the intervals

y \leq 9 and - 1 \leq y \leq 1

Merge Overlapping Intervals

y \leq 9 and - 1 \leq y \leq 1

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

y \leq 9

and

- 1 \leq y \leq 1

- 1 \leq y \leq 1

- 1 \leq y \leq 1

True for all x

True for all x \in R

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