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(9sin(t)+1)/8 <= 2pi

Solution

\frac{9 sin ( t ) + 1}{8} \leq 2 π

True for all t \in R

Interval Notation

(- \infty, \infty)

Solution steps

\frac{9 sin ( t ) + 1}{8} \leq 2 π

Multiply both sides by

8

\frac{9 sin ( t ) + 1}{8} \leq 2 π

Multiply both sides by

8

\frac{8 ( 9 sin ( t ) + 1 )}{8} \leq 8 \cdot 2 π

Simplify

9 sin (t) + 1 \leq 16 π

9 sin (t) + 1 \leq 16 π

Move

1

to the right side

9 sin (t) + 1 \leq 16 π

Subtract

1

from both sides

9 sin (t) + 1 - 1 \leq 16 π - 1

Simplify

9 sin (t) \leq 16 π - 1

9 sin (t) \leq 16 π - 1

Divide both sides by

9

9 sin (t) \leq 16 π - 1

Divide both sides by

9

\frac{9 sin ( t )}{9} \leq \frac{16 π}{9} - \frac{1}{9}

Simplify

\frac{9 sin ( t )}{9} \leq \frac{16 π}{9} - \frac{1}{9}

Simplify

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

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

Divide the numbers:

\frac{9}{9} = 1

= sin (t)

Simplify

\frac{16 π}{9} - \frac{1}{9} : \frac{16 π - 1}{9}

\frac{16 π}{9} - \frac{1}{9}

Apply rule

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

= \frac{16 π - 1}{9}

sin (t) \leq \frac{16 π - 1}{9}

sin (t) \leq \frac{16 π - 1}{9}

sin (t) \leq \frac{16 π - 1}{9}

Range of

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

Function range definition

The range of the basic

sin

function is

- 1 \leq sin (t) \leq 1

- 1 \leq sin (t) \leq 1

sin (t) \leq \frac{16 π - 1}{9} and - 1 \leq sin (t) \leq 1 : - 1 \leq sin (t) \leq 1

Let

y = sin (t)

Combine the intervals

y \leq \frac{16 π - 1}{9} and - 1 \leq y \leq 1

Merge Overlapping Intervals

y \leq \frac{16 π - 1}{9} and - 1 \leq y \leq 1

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

y \leq \frac{16 π - 1}{9}

and

- 1 \leq y \leq 1

- 1 \leq y \leq 1

- 1 \leq y \leq 1

True for all t

True for all t \in R