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Popular >

8-9sin(x)cos(x)>20

Solution

8 - 9 sin (x) cos (x) > 20

Solution

False for all x \in R

Solution steps

8 - 9 sin (x) cos (x) > 20

Use the following identity:

2 cos (x) sin (x) = sin (2 x)

Therefore

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

8 - 9 \cdot \frac{sin ( 2 x )}{2} > 20

Simplify

8 - 9 \cdot \frac{sin ( 2 x )}{2} : 8 - \frac{9}{2} sin (2 x)

8 - 9 \cdot \frac{sin ( 2 x )}{2}

Multiply

9 \cdot \frac{sin ( 2 x )}{2} : \frac{9 sin ( 2 x )}{2}

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

Multiply fractions:

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

= \frac{sin ( 2 x ) \cdot 9}{2}

= 8 - \frac{9 sin ( 2 x )}{2}

= 8 - \frac{9}{2} sin (2 x)

8 - \frac{9}{2} sin (2 x) > 20

Move

8

to the right side

8 - \frac{9}{2} sin (2 x) > 20

Subtract

8

from both sides

8 - \frac{9}{2} sin (2 x) - 8 > 20 - 8

Simplify

- \frac{9}{2} sin (2 x) > 12

- \frac{9}{2} sin (2 x) > 12

Multiply both sides by

- 1

- \frac{9}{2} sin (2 x) > 12

Multiply both sides by -1 (reverse the inequality)

(- \frac{9}{2} sin (2 x)) (- 1) < 12 (- 1)

Simplify

\frac{9}{2} sin (2 x) < - 12

\frac{9}{2} sin (2 x) < - 12

Multiply both sides by

2

\frac{9}{2} sin (2 x) < - 12

Multiply both sides by

2

2 \cdot \frac{9}{2} sin (2 x) < 2 (- 12)

Simplify

9 sin (2 x) < - 24

9 sin (2 x) < - 24

Divide both sides by

9

9 sin (2 x) < - 24

Divide both sides by

9

\frac{9 sin ( 2 x )}{9} < \frac{- 24}{9}

Simplify

\frac{9 sin ( 2 x )}{9} < \frac{- 24}{9}

Simplify

\frac{9 sin ( 2 x )}{9} : sin (2 x)

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

Divide the numbers:

\frac{9}{9} = 1

= sin (2 x)

Simplify

\frac{- 24}{9} : - \frac{8}{3}

\frac{- 24}{9}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{24}{9}

Cancel the common factor:

3

= - \frac{8}{3}

sin (2 x) < - \frac{8}{3}

sin (2 x) < - \frac{8}{3}

sin (2 x) < - \frac{8}{3}

Range of

sin (2 x) : - 1 \leq sin (2 x) \leq 1

Function range definition

The range of the basic

sin

function is

- 1 \leq sin (2 x) \leq 1

- 1 \leq sin (2 x) \leq 1

sin (2 x) < - \frac{8}{3} and - 1 \leq sin (2 x) \leq 1 :

False

Let

y = sin (2 x)

Combine the intervals

y < - \frac{8}{3} and - 1 \leq y \leq 1

Merge Overlapping Intervals

y < - \frac{8}{3} and - 1 \leq y \leq 1

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

y < - \frac{8}{3}

and

- 1 \leq y \leq 1

False for all y \in R

False for all y \in R

No Solution for x \in R

False for all x \in R

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