English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Trigonometry >

(sin(x))/(cos(x))>= 2sin(x)*cos(x)

Solution

\frac{sin ( x )}{cos ( x )} \geq 2 sin (x) \cdot cos (x)

Solution

\frac{π}{4} + πn \leq x < \frac{π}{2} + πn or \frac{3 π}{4} + πn \leq x \leq π + πn

Interval Notation

[\frac{π}{4} + πn, \frac{π}{2} + πn) \cup [\frac{3 π}{4} + πn, π + πn]

Decimal

0.78539 \dots + πn \leq x < 1.57079 \dots + πn or 2.35619 \dots + πn \leq x \leq 3.14159 \dots + πn

Solution steps

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

Move

2 sin (x) cos (x)

to the left side

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

Subtract

2 sin (x) cos (x)

from both sides

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

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

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

Simplify

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

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

Convert element to fraction:

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

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

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

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

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

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

Apply exponent rule:

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

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

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

Add the numbers:

1 + 1 = 2

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

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

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

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

Periodicity of

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

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

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

Periodicity of

\frac{sin ( x )}{cos ( x )} : π

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

is composed of the following functions and periods:

sin (x)

with periodicity of

2 π

The compound periodicity is:

π

Periodicity of

2 sin (x) cos (x) : π

2 sin (x) cos (x)

is composed of the following functions and periods:

sin (x)

with periodicity of

2 π

The compound periodicity is:

π

Combine periods:

π, π

= π

Find the zeroes and undifined points of

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

for

0 \leq x < π

To find the zeroes, set the inequality to zero

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

\frac{sin ( x ) - 2 cos ^{2} ( x ) sin ( x )}{cos ( x )} = 0, 0 \leq x < π : x = 0, x = \frac{3 π}{4}, x = \frac{π}{4}

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

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

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

Factor

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

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

Factor out common term

- sin (x)

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

Factor

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

2 cos^{2} (x) - 1

Rewrite

2 cos^{2} (x) - 1

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

2 cos^{2} (x) - 1

Apply radical rule:

a = (a)^{2}

2 = (2)^{2}

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

Rewrite

1

1^{2}

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

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

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

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

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

Apply Difference of Two Squares Formula:

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

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

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

= - sin (x) (2 cos (x) + 1) (2 cos (x) - 1)

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

Solving each part separately

sin (x) = 0 or 2 cos (x) + 1 = 0 or 2 cos (x) - 1 = 0

sin (x) = 0, 0 \leq x < π : x = 0

sin (x) = 0, 0 \leq x < 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 < π

x = 0

2 cos (x) + 1 = 0, 0 \leq x < π : x = \frac{3 π}{4}

2 cos (x) + 1 = 0, 0 \leq x < π

Move

1

to the right side

2 cos (x) + 1 = 0

Subtract

1

from both sides

2 cos (x) + 1 - 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

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

Simplify

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

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

Cancel the common factor:

2

= cos (x)

Simplify

\frac{- 1}{2} : - \frac{2}{2}

\frac{- 1}{2}

Apply the fraction rule:

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

= - \frac{1}{2}

Rationalize

- \frac{1}{2} : - \frac{2}{2}

- \frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

= - \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2}

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

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

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

General solutions for

cos (x) = - \frac{2}{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{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

Solutions for the range

0 \leq x < π

x = \frac{3 π}{4}

2 cos (x) - 1 = 0, 0 \leq x < π : x = \frac{π}{4}

2 cos (x) - 1 = 0, 0 \leq x < π

Move

1

to the right side

2 cos (x) - 1 = 0

Add

1

to both sides

2 cos (x) - 1 + 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

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

Simplify

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

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

Cancel the common factor:

2

= cos (x)

Simplify

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

= \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2}{2}

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

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

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

General solutions for

cos (x) = \frac{2}{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}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Solutions for the range

0 \leq x < π

x = \frac{π}{4}

Combine all the solutions

x = 0, x = \frac{3 π}{4}, x = \frac{π}{4}

Find the undefined points:

x = \frac{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 < π

x = \frac{π}{2}

0, \frac{π}{4}, \frac{π}{2}, \frac{3 π}{4}

Identify the intervals

0 < x < \frac{π}{4}, \frac{π}{4} < x < \frac{π}{2}, \frac{π}{2} < x < \frac{3 π}{4}, \frac{3 π}{4} < x < π

Summarize in a table:

sin (x) - 2 cos^{2} (x) sin (x) cos (x) \frac{s i n ( x ) - 2 c o s ^{2} ( x ) s i n ( x )}{c o s ( x )} x = 0 0 + 0 0 < x < \frac{π}{4} - + - x = \frac{π}{4} 0 + 0 \frac{π}{4} < x < \frac{π}{2} + + + x = \frac{π}{2} + 0 Undefined \frac{π}{2} < x < \frac{3 π}{4} + - - x = \frac{3 π}{4} 0 - 0 \frac{3 π}{4} < x < π - - + x = π 0 - 0

Identify the intervals that satisfy the required condition:

\geq 0

x = 0 or x = \frac{π}{4} or \frac{π}{4} < x < \frac{π}{2} or x = \frac{3 π}{4} or \frac{3 π}{4} < x < π or x = π

Merge Overlapping Intervals

x = 0 or \frac{π}{4} \leq x < \frac{π}{2} or \frac{3 π}{4} \leq x < π or x = π