English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tan(x)tan(2x)>1

Solution

tan (x) tan (2 x) > 1

Solution

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

Interval Notation

(πn, \frac{π}{4} + πn) \cup (\frac{3 π}{4} + πn, π + πn)

Decimal Notation

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

Solution steps

tan (x) tan (2 x) > 1

Periodicity of

tan (x) tan (2 x) : π

tan (x) tan (2 x)

is composed of the following functions and periods:

tan (x)

with periodicity of

π

The compound periodicity is:

= π

Express with sin, cos

tan (x) tan (2 x) > 1

Use the basic trigonometric identity:

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

\frac{sin ( x )}{cos ( x )} tan (2 x) > 1

Use the basic trigonometric identity:

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

\frac{sin ( x )}{cos ( x )} \cdot \frac{sin ( 2 x )}{cos ( 2 x )} > 1

\frac{sin ( x )}{cos ( x )} \cdot \frac{sin ( 2 x )}{cos ( 2 x )} > 1

Simplify

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

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

Multiply fractions:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

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

\frac{sin ( x ) sin ( 2 x )}{cos ( x ) cos ( 2 x )} > 1

Find the zeroes and undifined points of

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

for

0 \leq x < π

To find the zeroes, set the inequality to zero

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

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

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

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

sin (x) sin (2 x) = 0

Solving each part separately

sin (x) = 0 or sin (2 x) = 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

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

sin (2 x) = 0, 0 \leq x < π

General solutions for

sin (2 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}

2 x = 0 + 2 πn, 2 x = π + 2 πn

2 x = 0 + 2 πn, 2 x = π + 2 πn

Solve

2 x = 0 + 2 πn : x = πn

2 x = 0 + 2 πn

0 + 2 πn = 2 πn

2 x = 2 πn

Divide both sides by

2

2 x = 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{2 πn}{2}

Simplify

x = πn

x = πn

Solve

2 x = π + 2 πn : x = \frac{π}{2} + πn

2 x = π + 2 πn

Divide both sides by

2

2 x = π + 2 πn

Divide both sides by

2

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

Simplify

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

x = πn, x = \frac{π}{2} + πn

Solutions for the range

0 \leq x < π

x = 0, x = \frac{π}{2}

Combine all the solutions

x = 0, x = \frac{π}{2}

Since the equation is undefined for:

\frac{π}{2}

x = 0

Find the undefined points:

x = \frac{π}{2}, x = \frac{π}{4}, x = \frac{3 π}{4}

Find the zeros of the denominator

cos (x) cos (2 x) = 0

Solving each part separately

cos (x) = 0 or cos (2 x) = 0

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

cos (x) = 0, 0 \leq x < π

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}

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

cos (2 x) = 0, 0 \leq x < π

General solutions for

cos (2 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}

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

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

Solve

2 x = \frac{π}{2} + 2 πn : x = \frac{π}{4} + πn

2 x = \frac{π}{2} + 2 πn

Divide both sides by

2

2 x = \frac{π}{2} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2} : \frac{π}{4} + πn

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{4} + πn

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

Solve

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

2 x = \frac{3 π}{2} + 2 πn

Divide both sides by

2

2 x = \frac{3 π}{2} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2} : \frac{3 π}{4} + πn

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{3 π}{2}}{2} = \frac{3 π}{4}

\frac{\frac{3 π}{2}}{2}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

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

Solutions for the range

0 \leq x < π

x = \frac{π}{4}, x = \frac{3 π}{4}

Combine all the solutions

x = \frac{π}{2}, x = \frac{π}{4}, x = \frac{3 π}{4}

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 (2 x) \frac{s i n ( x ) s i n ( 2 x )}{c o s ( x ) c o s ( 2 x )} x = 0 00 + + 0 0 < x < \frac{π}{4} + + + + + x = \frac{π}{4} + + + 0 Undefined \frac{π}{4} < x < \frac{π}{2} + + + - - x = \frac{π}{2} + 00 - Undefined \frac{π}{2} < x < \frac{3 π}{4} + - - - - x = \frac{3 π}{4} + - - 0 Undefined \frac{3 π}{4} < x < π + - - + + x = π 00 - + 0

Identify the intervals that satisfy the required condition:

> 0

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

Apply the periodicity of

tan (x) tan (2 x)

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

Popular Examples

tan(a)<0

tan (a) < 0

1-sin(x)>= 0

1 - sin (x) \geq 0

5/4 sin(3/4 x-6)+1/4 <= 3/2

\frac{5}{4} sin (\frac{3}{4} x - 6) + \frac{1}{4} \leq \frac{3}{2}

-1/2 (10sin(2pi*60x))>-0.52

- \frac{1}{2} (10 sin (2 π \cdot 60 x)) > - 0.52

sin(t)>= 0

sin (t) \geq 0