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Popular >

tan(x)>= sin(x)

Solution

tan (x) \geq sin (x)

Solution

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

Interval Notation

[2 πn, \frac{π}{2} + 2 πn) \cup [π + 2 πn, \frac{3 π}{2} + 2 πn)

Decimal Notation

2 πn \leq x < 1.57079 \dots + 2 πn or 3.14159 \dots + 2 πn \leq x < 4.71238 \dots + 2 πn

Solution steps

tan (x) \geq sin (x)

Move

sin (x)

to the left side

tan (x) \geq sin (x)

Subtract

sin (x)

from both sides

tan (x) - sin (x) \geq sin (x) - sin (x)

tan (x) - sin (x) \geq 0

tan (x) - sin (x) \geq 0

Periodicity of

tan (x) - sin (x) : 2 π

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

tan (x), sin (x)

Periodicity of

tan (x) : π

Periodicity of

tan (x)

π

= π

Periodicity of

sin (x) : 2 π

Periodicity of

sin (x)

2 π

= 2 π

Combine periods:

π, 2 π

= 2 π

Express with sin, cos

tan (x) - sin (x) \geq 0

Use the basic trigonometric identity:

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

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

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

Find the zeroes and undifined points of

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

for

0 \leq x < 2 π

To find the zeroes, set the inequality to zero

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

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

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

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

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

Factor

sin (x) - sin (x) cos (x) : - sin (x) (cos (x) - 1)

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

Factor out common term

- sin (x)

= - sin (x) (- 1 + cos (x))

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

Solving each part separately

sin (x) = 0 or cos (x) - 1 = 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) - 1 = 0, 0 \leq x < 2 π : x = 0

cos (x) - 1 = 0, 0 \leq x < 2 π

Move

1

to the right side

cos (x) - 1 = 0

Add

1

to both sides

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

Simplify

cos (x) = 1

cos (x) = 1

General solutions for

cos (x) = 1

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 = 0 + 2 πn

x = 0 + 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

Solutions for the range

0 \leq x < 2 π

x = 0

Combine all the solutions

x = 0, x = π

Find the undefined points:

x = \frac{π}{2}, x = \frac{3 π}{2}

Find the zeros of the denominator

cos (x) = 0

General solutions for

cos (x) = 0

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}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

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

Solutions for the range

0 \leq x < 2 π

x = \frac{π}{2}, x = \frac{3 π}{2}

0, \frac{π}{2}, π, \frac{3 π}{2}

Identify the intervals

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

Summarize in a table:

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

Identify the intervals that satisfy the required condition:

\geq 0

x = 0 or 0 < x < \frac{π}{2} or x = π or π < x < \frac{3 π}{2} or x = 2 π

Merge Overlapping Intervals

0 \leq x < \frac{π}{2} or π \leq x < \frac{3 π}{2} or x = 2 π

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

x = 0

0 < x < \frac{π}{2}

0 \leq x < \frac{π}{2}

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

0 \leq x < \frac{π}{2}

x = π

0 \leq x < \frac{π}{2} or x = π

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

0 \leq x < \frac{π}{2} or x = π

π < x < \frac{3 π}{2}

0 \leq x < \frac{π}{2} or π \leq x < \frac{3 π}{2}

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

0 \leq x < \frac{π}{2} or π \leq x < \frac{3 π}{2}

x = 2 π

0 \leq x < \frac{π}{2} or π \leq x < \frac{3 π}{2} or x = 2 π

0 \leq x < \frac{π}{2} or π \leq x < \frac{3 π}{2} or x = 2 π

Apply the periodicity of

tan (x) - sin (x)

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

Popular Examples

1<= tan(x)

1 \leq tan (x)

tan(x)<-(sqrt(3))/3

tan (x) < - \frac{3}{3}

sin(9x^2)>0

sin (9 x^{2}) > 0

2cos(2x)-1>0

2 cos (2 x) - 1 > 0

sin(-pi/2 x-pi/2)>-15

sin (- \frac{π}{2} x - \frac{π}{2}) > - 15