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Popular >

cot(x)+(sin(x))/(cos(x)-2)>= 0

Solution

cot (x) + \frac{sin ( x )}{cos ( x ) - 2} \geq 0

Solution

2 πn < x \leq \frac{π}{3} + 2 πn or π + 2 πn < x \leq \frac{5 π}{3} + 2 πn

Interval Notation

(2 πn, \frac{π}{3} + 2 πn] \cup (π + 2 πn, \frac{5 π}{3} + 2 πn]

Decimal Notation

2 πn < x \leq 1.04719 \dots + 2 πn or 3.14159 \dots + 2 πn < x \leq 5.23598 \dots + 2 πn

Solution steps

cot (x) + \frac{sin ( x )}{cos ( x ) - 2} \geq 0

Periodicity of

cot (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

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

Periodicity of

cot (x) : π

Periodicity of

cot (x)

π

= 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:

cot (x)

with periodicity of

π

The compound periodicity is:

2 π

Combine periods:

π, 2 π

= 2 π

Express with sin, cos

cot (x) + \frac{sin ( x )}{cos ( x ) - 2} \geq 0

Use the basic trigonometric identity:

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

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

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

Simplify

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

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

Least Common Multiplier of

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

sin (x), cos (x) - 2

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

sin (x)

cos (x) - 2

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

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

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

For

\frac{cos ( x )}{sin ( x )} :

multiply the denominator and numerator by

cos (x) - 2

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

For

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

multiply the denominator and numerator by

sin (x)

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

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

\frac{cos ( x ) ( cos ( x ) - 2 ) + sin ^{2} ( x )}{sin ( x ) ( cos ( x ) - 2 )} \geq 0

Find the zeroes and undifined points of

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

for

0 \leq x < 2 π

To find the zeroes, set the inequality to zero

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

\frac{cos ( x ) ( cos ( x ) - 2 ) + sin ^{2} ( x )}{sin ( x ) ( cos ( x ) - 2 )} = 0, 0 \leq x < 2 π : x = \frac{π}{3}, x = \frac{5 π}{3}

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

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

cos (x) (cos (x) - 2) + sin^{2} (x) = 0

Rewrite using trig identities

sin^{2} (x) + (- 2 + cos (x)) cos (x)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= 1 - cos^{2} (x) + (- 2 + cos (x)) cos (x)

Simplify

1 - cos^{2} (x) + (- 2 + cos (x)) cos (x) : - 2 cos (x) + 1

1 - cos^{2} (x) + (- 2 + cos (x)) cos (x)

= 1 - cos^{2} (x) + cos (x) (- 2 + cos (x))

Expand

cos (x) (- 2 + cos (x)) : - 2 cos (x) + cos^{2} (x)

cos (x) (- 2 + cos (x))

Apply the distributive law:

a (b + c) = ab + a c

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

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

Apply minus-plus rules

+ (- a) = - a

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

cos (x) cos (x) = cos^{2} (x)

cos (x) cos (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos (x) cos (x) = cos^{1 + 1} (x)

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

= - 2 cos (x) + cos^{2} (x)

= 1 - cos^{2} (x) - 2 cos (x) + cos^{2} (x)

Simplify

1 - cos^{2} (x) - 2 cos (x) + cos^{2} (x) : - 2 cos (x) + 1

1 - cos^{2} (x) - 2 cos (x) + cos^{2} (x)

Group like terms

= - cos^{2} (x) - 2 cos (x) + cos^{2} (x) + 1

Add similar elements:

- cos^{2} (x) + cos^{2} (x) = 0

= - 2 cos (x) + 1

= - 2 cos (x) + 1

= - 2 cos (x) + 1

1 - 2 cos (x) = 0

Move

1

to the right side

1 - 2 cos (x) = 0

Subtract

1

from both sides

1 - 2 cos (x) - 1 = 0 - 1

Simplify

- 2 cos (x) = - 1

- 2 cos (x) = - 1

Divide both sides by

- 2

- 2 cos (x) = - 1

Divide both sides by

- 2

\frac{- 2 cos ( x )}{- 2} = \frac{- 1}{- 2}

Simplify

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

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

General solutions for

cos (x) = \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}

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Solutions for the range

0 \leq x < 2 π

x = \frac{π}{3}, x = \frac{5 π}{3}

Find the undefined points:

x = 0, x = π

Find the zeros of the denominator

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

Solving each part separately

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

No Solution

cos (x) - 2 = 0, 0 \leq x < 2 π

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

Combine all the solutions

x = 0, x = π

0, \frac{π}{3}, π, \frac{5 π}{3}

Identify the intervals

0 < x < \frac{π}{3}, \frac{π}{3} < x < π, π < x < \frac{5 π}{3}, \frac{5 π}{3} < x < 2 π

Summarize in a table:

cos (x) (cos (x) - 2) + sin^{2} (x) sin (x) cos (x) - 2 \frac{c o s ( x ) ( c o s ( x ) - 2 ) + s i n ^{2} ( x )}{s i n ( x ) ( c o s ( x ) - 2 )} x = 0 - 0 - Undefined 0 < x < \frac{π}{3} - + - + x = \frac{π}{3} 0 + - 0 \frac{π}{3} < x < π + + - - x = π + 0 - Undefined π < x < \frac{5 π}{3} + - - + x = \frac{5 π}{3} 0 - - 0 \frac{5 π}{3} < x < 2 π - - - - x = 2 π - 0 - Undefined

Identify the intervals that satisfy the required condition:

\geq 0

0 < x < \frac{π}{3} or x = \frac{π}{3} or π < x < \frac{5 π}{3} or x = \frac{5 π}{3}

Merge Overlapping Intervals

0 < x \leq \frac{π}{3} or π < x < \frac{5 π}{3} or x = \frac{5 π}{3}

The union of two intervals is the set of numbers which are in either interval

0 < x < \frac{π}{3}

x = \frac{π}{3}

0 < x \leq \frac{π}{3}

The union of two intervals is the set of numbers which are in either interval

0 < x \leq \frac{π}{3}

π < x < \frac{5 π}{3}

0 < x \leq \frac{π}{3} or π < x < \frac{5 π}{3}

The union of two intervals is the set of numbers which are in either interval

0 < x \leq \frac{π}{3} or π < x < \frac{5 π}{3}

x = \frac{5 π}{3}

0 < x \leq \frac{π}{3} or π < x \leq \frac{5 π}{3}

0 < x \leq \frac{π}{3} or π < x \leq \frac{5 π}{3}

Apply the periodicity of

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

2 πn < x \leq \frac{π}{3} + 2 πn or π + 2 πn < x \leq \frac{5 π}{3} + 2 πn

Popular Examples

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cos(x)<=-(sqrt(2))/2 ,-pi<= x<= pi

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6sin(2x-(2pi)/3)>0

6 sin (2 x - \frac{2 π}{3}) > 0

1>tan(x)

1 > tan (x)

cos(x)-(sqrt(3))/2 <= 0

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