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Popular >

sin(x)>(sin(x))/(cos(x)-2)

Solution

sin (x) > \frac{sin ( x )}{cos ( x ) - 2}

Solution

2 πn < x < π + 2 πn

Interval Notation

(2 πn, π + 2 πn)

Decimal Notation

2 πn < x < 3.14159 \dots + 2 πn

Solution steps

sin (x) > \frac{sin ( x )}{cos ( x ) - 2}

Move

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

to the left side

sin (x) > \frac{sin ( x )}{cos ( x ) - 2}

Subtract

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

from both sides

sin (x) - \frac{sin ( x )}{cos ( x ) - 2} > \frac{sin ( x )}{cos ( x ) - 2} - \frac{sin ( x )}{cos ( x ) - 2}

sin (x) - \frac{sin ( x )}{cos ( x ) - 2} > 0

sin (x) - \frac{sin ( x )}{cos ( x ) - 2} > 0

Simplify

sin (x) - \frac{sin ( x )}{cos ( x ) - 2} : \frac{sin ( x ) cos ( x ) - 3 sin ( x )}{cos ( x ) - 2}

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Expand

sin (x) (cos (x) - 2) - sin (x) : sin (x) cos (x) - 3 sin (x)

sin (x) (cos (x) - 2) - sin (x)

Expand

sin (x) (cos (x) - 2) : sin (x) cos (x) - 2 sin (x)

sin (x) (cos (x) - 2)

Apply the distributive law:

a (b - c) = ab - a c

a = sin (x), b = cos (x), c = 2

= sin (x) cos (x) - sin (x) \cdot 2

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

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

Add similar elements:

- 2 sin (x) - sin (x) = - 3 sin (x)

= sin (x) cos (x) - 3 sin (x)

= \frac{sin ( x ) cos ( x ) - 3 sin ( x )}{cos ( x ) - 2}

\frac{sin ( x ) cos ( x ) - 3 sin ( x )}{cos ( x ) - 2} > 0

Periodicity of

sin (x) - \frac{sin ( x )}{cos ( x ) - 2} : 2 π

The compound periodicity of the sum of periodic functions is the least common multiplier of the periods

sin (x), \frac{sin ( x )}{cos ( x ) - 2}

Periodicity of

sin (x) : 2 π

Periodicity of

sin (x)

2 π

= 2 π

Periodicity of

\frac{sin ( x )}{cos ( x ) - 2} : 2 π

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

is composed of the following functions and periods:

sin (x)

with periodicity of

2 π

The compound periodicity is:

2 π

Combine periods:

2 π, 2 π

= 2 π

Find the zeroes and undifined points of

\frac{sin ( x ) cos ( x ) - 3 sin ( x )}{cos ( x ) - 2}

for

0 \leq x < 2 π

To find the zeroes, set the inequality to zero

\frac{sin ( x ) cos ( x ) - 3 sin ( x )}{cos ( x ) - 2} = 0

\frac{sin ( x ) cos ( x ) - 3 sin ( x )}{cos ( x ) - 2} = 0, 0 \leq x < 2 π : x = 0, x = π

\frac{sin ( x ) cos ( x ) - 3 sin ( x )}{cos ( x ) - 2} = 0, 0 \leq x < 2 π

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

sin (x) cos (x) - 3 sin (x) = 0

Factor

sin (x) cos (x) - 3 sin (x) : sin (x) (cos (x) - 3)

sin (x) cos (x) - 3 sin (x)

Factor out common term

sin (x)

= sin (x) (cos (x) - 3)

sin (x) (cos (x) - 3) = 0

Solving each part separately

sin (x) = 0 or cos (x) - 3 = 0

sin (x) = 0, 0 \leq x < 2 π : x = 0, x = π

sin (x) = 0, 0 \leq x < 2 π

General solutions for

sin (x) = 0

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

Solutions for the range

0 \leq x < 2 π

x = 0, x = π

cos (x) - 3 = 0, 0 \leq x < 2 π :

No Solution

cos (x) - 3 = 0, 0 \leq x < 2 π

Move

3

to the right side

cos (x) - 3 = 0

Add

3

to both sides

cos (x) - 3 + 3 = 0 + 3

Simplify

cos (x) = 3

cos (x) = 3

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = 0, x = π

Find the undefined points:

No Solution

Find the zeros of the denominator

cos (x) - 2 = 0

Move

2

to the right side

cos (x) - 2 = 0

Add

2

to both sides

cos (x) - 2 + 2 = 0 + 2

Simplify

cos (x) = 2

cos (x) = 2

- 1 \leq cos (x) \leq 1

No Solution for x \in R

0, π

Identify the intervals

0 < x < π, π < x < 2 π

Summarize in a table:

sin (x) cos (x) - 3 sin (x) cos (x) - 2 \frac{s i n ( x ) c o s ( x ) - 3 s i n ( x )}{c o s ( x ) - 2} x = 0 0 - 0 0 < x < π - - + x = π 0 - 0 π < x < 2 π + - - x = 2 π 0 - 0

Identify the intervals that satisfy the required condition:

> 0

0 < x < π

Apply the periodicity of

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

2 πn < x < π + 2 πn

Popular Examples

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cot (x) \leq 3

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2 cos (x) \geq 1

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