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Popular >

2tan(2x)<= 3tan(x)

Solution

2 tan (2 x) \leq 3 tan (x)

Solution

\frac{π}{4} + πn < x < \frac{π}{2} + πn or \frac{3 π}{4} + πn < 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 < x < 1.57079 \dots + πn or 2.35619 \dots + πn < x \leq 3.14159 \dots + πn

Solution steps

2 tan (2 x) \leq 3 tan (x)

Move

3 tan (x)

to the left side

2 tan (2 x) \leq 3 tan (x)

Subtract

3 tan (x)

from both sides

2 tan (2 x) - 3 tan (x) \leq 3 tan (x) - 3 tan (x)

2 tan (2 x) - 3 tan (x) \leq 0

2 tan (2 x) - 3 tan (x) \leq 0

Periodicity of

2 tan (2 x) - 3 tan (x) : π

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

2 tan (2 x), 3 tan (x)

Periodicity of

2 tan (2 x) : \frac{π}{2}

Periodicity of

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

Periodicity of

tan (x)

π

= \frac{π}{∣ 2 ∣}

Simplify

= \frac{π}{2}

Periodicity of

3 tan (x) : π

Periodicity of

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

Periodicity of

tan (x)

π

= \frac{π}{∣ 1 ∣}

Simplify

= π

Combine periods:

\frac{π}{2}, π

= π

Express with sin, cos

2 tan (2 x) - 3 tan (x) \leq 0

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

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

Simplify

2 \cdot \frac{sin ( 2 x )}{cos ( 2 x )} - 3 \cdot \frac{sin ( x )}{cos ( x )} : \frac{2 sin ( 2 x ) cos ( x ) - 3 sin ( x ) cos ( 2 x )}{cos ( 2 x ) cos ( x )}

2 \cdot \frac{sin ( 2 x )}{cos ( 2 x )} - 3 \cdot \frac{sin ( x )}{cos ( x )}

Multiply

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

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

Multiply fractions:

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

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

= \frac{2 sin ( 2 x )}{cos ( 2 x )} - 3 \cdot \frac{sin ( x )}{cos ( x )}

Multiply

3 \cdot \frac{sin ( x )}{cos ( x )} : \frac{3 sin ( x )}{cos ( x )}

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

Multiply fractions:

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

= \frac{sin ( x ) \cdot 3}{cos ( x )}

= \frac{2 sin ( 2 x )}{cos ( 2 x )} - \frac{3 sin ( x )}{cos ( x )}

Least Common Multiplier of

cos (2 x), cos (x) : cos (2 x) cos (x)

cos (2 x), cos (x)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

cos (2 x)

cos (x)

= cos (2 x) cos (x)

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

cos (2 x) cos (x)

For

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

multiply the denominator and numerator by

cos (x)

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

For

\frac{sin ( x ) \cdot 3}{cos ( x )} :

multiply the denominator and numerator by

cos (2 x)

\frac{sin ( x ) \cdot 3}{cos ( x )} = \frac{sin ( x ) \cdot 3 cos ( 2 x )}{cos ( x ) cos ( 2 x )}

= \frac{sin ( 2 x ) \cdot 2 cos ( x )}{cos ( 2 x ) cos ( x )} - \frac{sin ( x ) \cdot 3 cos ( 2 x )}{cos ( x ) cos ( 2 x )}

Since the denominators are equal, combine the fractions:

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

= \frac{sin ( 2 x ) \cdot 2 cos ( x ) - sin ( x ) \cdot 3 cos ( 2 x )}{cos ( 2 x ) cos ( x )}

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

Find the zeroes and undifined points of

\frac{2 sin ( 2 x ) cos ( x ) - 3 sin ( x ) cos ( 2 x )}{cos ( 2 x ) cos ( x )}

for

0 \leq x < π

To find the zeroes, set the inequality to zero

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

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

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

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

2 sin (2 x) cos (x) - 3 sin (x) cos (2 x) = 0

Rewrite using trig identities

2 cos (x) sin (2 x) - 3 cos (2 x) sin (x)

Use the Double Angle identity:

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

= 2 cos (x) \cdot 2 sin (x) cos (x) - 3 cos (2 x) sin (x)

2 cos (x) \cdot 2 sin (x) cos (x) = 4 cos^{2} (x) sin (x)

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

Multiply the numbers:

2 \cdot 2 = 4

= 4 cos (x) sin (x) cos (x)

Apply exponent rule:

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

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

= 4 sin (x) cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= 4 sin (x) cos^{2} (x)

= 4 cos^{2} (x) sin (x) - 3 cos (2 x) sin (x)

- 3 cos (2 x) sin (x) + 4 cos^{2} (x) sin (x) = 0

Factor

- 3 cos (2 x) sin (x) + 4 cos^{2} (x) sin (x) : sin (x) (- 3 cos (2 x) + 4 cos^{2} (x))

- 3 cos (2 x) sin (x) + 4 cos^{2} (x) sin (x)

Factor out common term

sin (x)

= sin (x) (- 3 cos (2 x) + 4 cos^{2} (x))

sin (x) (- 3 cos (2 x) + 4 cos^{2} (x)) = 0

Solving each part separately

sin (x) = 0 or - 3 cos (2 x) + 4 cos^{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

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

No Solution

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

Rewrite using trig identities

- 3 cos (2 x) + 4 cos^{2} (x)

Use the Double Angle identity:

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

= - 3 (2 cos^{2} (x) - 1) + 4 cos^{2} (x)

Simplify

- 3 (2 cos^{2} (x) - 1) + 4 cos^{2} (x) : - 2 cos^{2} (x) + 3

- 3 (2 cos^{2} (x) - 1) + 4 cos^{2} (x)

Expand

- 3 (2 cos^{2} (x) - 1) : - 6 cos^{2} (x) + 3

- 3 (2 cos^{2} (x) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = - 3, b = 2 cos^{2} (x), c = 1

= - 3 \cdot 2 cos^{2} (x) - (- 3) \cdot 1

Apply minus-plus rules

- (- a) = a

= - 3 \cdot 2 cos^{2} (x) + 3 \cdot 1

Simplify

- 3 \cdot 2 cos^{2} (x) + 3 \cdot 1 : - 6 cos^{2} (x) + 3

- 3 \cdot 2 cos^{2} (x) + 3 \cdot 1

Multiply the numbers:

3 \cdot 2 = 6

= - 6 cos^{2} (x) + 3 \cdot 1

Multiply the numbers:

3 \cdot 1 = 3

= - 6 cos^{2} (x) + 3

= - 6 cos^{2} (x) + 3

= - 6 cos^{2} (x) + 3 + 4 cos^{2} (x)

Simplify

- 6 cos^{2} (x) + 3 + 4 cos^{2} (x) : - 2 cos^{2} (x) + 3

- 6 cos^{2} (x) + 3 + 4 cos^{2} (x)

Group like terms

= - 6 cos^{2} (x) + 4 cos^{2} (x) + 3

Add similar elements:

- 6 cos^{2} (x) + 4 cos^{2} (x) = - 2 cos^{2} (x)

= - 2 cos^{2} (x) + 3

= - 2 cos^{2} (x) + 3

= - 2 cos^{2} (x) + 3

3 - 2 cos^{2} (x) = 0

Solve by substitution

3 - 2 cos^{2} (x) = 0

Let:

cos (x) = u

3 - 2 u^{2} = 0

3 - 2 u^{2} = 0 : u = \frac{3}{2}, u = - \frac{3}{2}

3 - 2 u^{2} = 0

Move

3

to the right side

3 - 2 u^{2} = 0

Subtract

3

from both sides

3 - 2 u^{2} - 3 = 0 - 3

Simplify

- 2 u^{2} = - 3

- 2 u^{2} = - 3

Divide both sides by

- 2

- 2 u^{2} = - 3

Divide both sides by

- 2

\frac{- 2 u ^{2}}{- 2} = \frac{- 3}{- 2}

Simplify

u^{2} = \frac{3}{2}

u^{2} = \frac{3}{2}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = \frac{3}{2}, u = - \frac{3}{2}

Substitute back

u = cos (x)

cos (x) = \frac{3}{2}, cos (x) = - \frac{3}{2}

cos (x) = \frac{3}{2}, cos (x) = - \frac{3}{2}

cos (x) = \frac{3}{2}, 0 \leq x < π :

No Solution

cos (x) = \frac{3}{2}, 0 \leq x < π

- 1 \leq cos (x) \leq 1

No Solution

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

No Solution

cos (x) = - \frac{3}{2}, 0 \leq x < π

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

No Solution

Combine all the solutions

x = 0

Find the undefined points:

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

Find the zeros of the denominator

cos (2 x) cos (x) = 0

Solving each part separately

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

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}

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}

Combine all the solutions

x = \frac{π}{4}, x = \frac{3 π}{4}, 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:

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

Identify the intervals that satisfy the required condition:

\leq 0

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

Merge Overlapping Intervals

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

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

x = 0

\frac{π}{4} < x < \frac{π}{2}

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

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

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

\frac{3 π}{4} < x < π

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

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

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

x = π

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

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

Apply the periodicity of

2 tan (2 x) - 3 tan (x)

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

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