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tan(x)>= sin(2x)

Solution

tan (x) \geq sin (2 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 Notation

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

Solution steps

tan (x) \geq sin (2 x)

Move

sin (2 x)

to the left side

tan (x) \geq sin (2 x)

Subtract

sin (2 x)

from both sides

tan (x) - sin (2 x) \geq sin (2 x) - sin (2 x)

tan (x) - sin (2 x) \geq 0

tan (x) - sin (2 x) \geq 0

Periodicity of

tan (x) - sin (2 x) : π

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

tan (x), sin (2 x)

Periodicity of

tan (x) : π

Periodicity of

tan (x)

π

= π

Periodicity of

sin (2 x) : π

Periodicity of

a \cdot sin (b x + c) + d = \frac{periodicity of sin ( x )}{∣ b ∣}

Periodicity of

sin (x)

2 π

= \frac{2 π}{∣ 2 ∣}

Simplify

= π

Combine periods:

π, π

= π

Express with sin, cos

tan (x) - sin (2 x) \geq 0

Use the basic trigonometric identity:

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

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

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

Simplify

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

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

Convert element to fraction:

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

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

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

Find the zeroes and undifined points of

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

for

0 \leq x < π

To find the zeroes, set the inequality to zero

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

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

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

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

cos (x) \cdot 2 sin (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)

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) - sin (2 x) cos (x) cos (x) \frac{s i n ( x ) - s i n ( 2 x ) c o s ( 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