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Popular >

cos(2x)<cos(4x)

Solution

cos (2 x) < cos (4 x)

Solution

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

Interval Notation

(\frac{π}{3} + πn, \frac{2 π}{3} + πn)

Decimal Notation

1.04719 \dots + πn < x < 2.09439 \dots + πn

Solution steps

cos (2 x) < cos (4 x)

Move

cos (4 x)

to the left side

cos (2 x) < cos (4 x)

Subtract

cos (4 x)

from both sides

cos (2 x) - cos (4 x) < cos (4 x) - cos (4 x)

cos (2 x) - cos (4 x) < 0

cos (2 x) - cos (4 x) < 0

Let:

u = 2 x

cos (u) - cos (2 u) < 0

cos (u) - cos (2 u) < 0 : \frac{2 π}{3} + 2 πn < u < \frac{4 π}{3} + 2 πn

cos (u) - cos (2 u) < 0

Use the following identity:

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

- (- 1 + 2 cos^{2} (u)) + cos (u) < 0

Simplify

1 - 2 cos^{2} (u) + cos (u) < 0

Let:

v = cos (u)

1 - 2 v^{2} + v < 0

1 - 2 v^{2} + v < 0 : v < - \frac{1}{2} or v > 1

1 - 2 v^{2} + v < 0

Factor

1 - 2 v^{2} + v : - (2 v + 1) (v - 1)

1 - 2 v^{2} + v

Factor out common term

- 1

= - (2 v^{2} - v - 1)

Factor

2 v^{2} - v - 1 : (2 v + 1) (v - 1)

2 v^{2} - v - 1

Write in the standard form

a x^{2} + b x + c

= 2 v^{2} - v - 1

Break the expression into groups

2 v^{2} - v - 1

Definition

Factors of

2 : 1, 2

2

Divisors (Factors)

Find the Prime factors of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Add 1

1

The factors of

2

1, 2

Negative factors of

2 : - 1, - 2

Multiply the factors by

- 1

to get the negative factors

- 1, - 2

For every two factors such that

u * v = - 2,

check if

u + v = - 1

Check

u = 1, v = - 2 : u * v = - 2, u + v = - 1 \Rightarrow

TrueCheck

u = 2, v = - 1 : u * v = - 2, u + v = 1 \Rightarrow

False

u = 1, v = - 2

Group into

(a x^{2} + ux) + (vx + c)

(2 v^{2} + v) + (- 2 v - 1)

= (2 v^{2} + v) + (- 2 v - 1)

Factor out

v

from

2 v^{2} + v : v (2 v + 1)

2 v^{2} + v

Apply exponent rule:

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

v^{2} = vv

= 2 vv + v

Factor out common term

v

= v (2 v + 1)

Factor out

- 1

from

- 2 v - 1 : - (2 v + 1)

- 2 v - 1

Factor out common term

- 1

= - (2 v + 1)

= v (2 v + 1) - (2 v + 1)

Factor out common term

2 v + 1

= (2 v + 1) (v - 1)

= - (2 v + 1) (v - 1)

- (2 v + 1) (v - 1) < 0

Multiply both sides by

- 1

(reverse the inequality)

(- (2 v + 1) (v - 1)) (- 1) > 0 \cdot (- 1)

Simplify

(2 v + 1) (v - 1) > 0

Identify the intervals

Find the signs of the factors of

(2 v + 1) (v - 1)

Find the signs of

2 v + 1

2 v + 1 = 0 : v = - \frac{1}{2}

2 v + 1 = 0

Move

1

to the right side

2 v + 1 = 0

Subtract

1

from both sides

2 v + 1 - 1 = 0 - 1

Simplify

2 v = - 1

2 v = - 1

Divide both sides by

2

2 v = - 1

Divide both sides by

2

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

Simplify

v = - \frac{1}{2}

v = - \frac{1}{2}

2 v + 1 < 0 : v < - \frac{1}{2}

2 v + 1 < 0

Move

1

to the right side

2 v + 1 < 0

Subtract

1

from both sides

2 v + 1 - 1 < 0 - 1

Simplify

2 v < - 1

2 v < - 1

Divide both sides by

2

2 v < - 1

Divide both sides by

2

\frac{2 v}{2} < \frac{- 1}{2}

Simplify

v < - \frac{1}{2}

v < - \frac{1}{2}

2 v + 1 > 0 : v > - \frac{1}{2}

2 v + 1 > 0

Move

1

to the right side

2 v + 1 > 0

Subtract

1

from both sides

2 v + 1 - 1 > 0 - 1

Simplify

2 v > - 1

2 v > - 1

Divide both sides by

2

2 v > - 1

Divide both sides by

2

\frac{2 v}{2} > \frac{- 1}{2}

Simplify

v > - \frac{1}{2}

v > - \frac{1}{2}

Find the signs of

v - 1

v - 1 = 0 : v = 1

v - 1 = 0

Move

1

to the right side

v - 1 = 0

Add

1

to both sides

v - 1 + 1 = 0 + 1

Simplify

v = 1

v = 1

v - 1 < 0 : v < 1

v - 1 < 0

Move

1

to the right side

v - 1 < 0

Add

1

to both sides

v - 1 + 1 < 0 + 1

Simplify

v < 1

v < 1

v - 1 > 0 : v > 1

v - 1 > 0

Move

1

to the right side

v - 1 > 0

Add

1

to both sides

v - 1 + 1 > 0 + 1

Simplify

v > 1

v > 1

Summarize in a table:

2 v + 1 v - 1 (2 v + 1) (v - 1) v < - \frac{1}{2} - - + v = - \frac{1}{2} 0 - 0 - \frac{1}{2} < v < 1 + - - v = 1 + 00 v > 1 + + +

Identify the intervals that satisfy the required condition:

> 0

v < - \frac{1}{2} or v > 1

v < - \frac{1}{2} or v > 1

v < - \frac{1}{2} or v > 1

Substitute back

v = cos (u)

cos (u) < - \frac{1}{2} or cos (u) > 1

cos (u) < - \frac{1}{2} : \frac{2 π}{3} + 2 πn < u < \frac{4 π}{3} + 2 πn

cos (u) < - \frac{1}{2}

For

cos (x) < a

, if

- 1 < a \leq 1

then

arccos (a) + 2 πn < x < 2 π - arccos (a) + 2 πn

arccos (- \frac{1}{2}) + 2 πn < u < 2 π - arccos (- \frac{1}{2}) + 2 πn

Simplify

arccos (- \frac{1}{2}) : \frac{2 π}{3}

arccos (- \frac{1}{2})

Use the following trivial identity:

arccos (- \frac{1}{2}) = \frac{2 π}{3}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{2 π}{3}

Simplify

2 π - arccos (- \frac{1}{2}) : \frac{4 π}{3}

2 π - arccos (- \frac{1}{2})

Use the following trivial identity:

arccos (- \frac{1}{2}) = \frac{2 π}{3}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= 2 π - \frac{2 π}{3}

Simplify

2 π - \frac{2 π}{3}

Convert element to fraction:

2 π = \frac{2 π 3}{3}

= \frac{2 π 3}{3} - \frac{2 π}{3}

Since the denominators are equal, combine the fractions:

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

= \frac{2 π 3 - 2 π}{3}

2 π 3 - 2 π = 4 π

2 π 3 - 2 π

Multiply the numbers:

2 \cdot 3 = 6

= 6 π - 2 π

Add similar elements:

6 π - 2 π = 4 π

= 4 π

= \frac{4 π}{3}

= \frac{4 π}{3}

\frac{2 π}{3} + 2 πn < u < \frac{4 π}{3} + 2 πn

cos (u) > 1 :

False for all

u \in R

cos (u) > 1

Range of

cos (u) : - 1 \leq cos (u) \leq 1

Function range definition

The range of the basic

cos

function is

- 1 \leq cos (u) \leq 1

- 1 \leq cos (u) \leq 1

cos (u) > 1 and - 1 \leq cos (u) \leq 1 :

False

Let

y = cos (u)

Combine the intervals

y > 1 and - 1 \leq y \leq 1

Merge Overlapping Intervals

y > 1 and - 1 \leq y \leq 1

The intersection of two intervals is the set of numbers which are in both intervals

y > 1

and

- 1 \leq y \leq 1

False for all y \in R

False for all y \in R

No Solution for u \in R

False for all u \in R

Combine the intervals

\frac{2 π}{3} + 2 πn < u < \frac{4 π}{3} + 2 πn or False for all u \in R

Merge Overlapping Intervals

\frac{2 π}{3} + 2 πn < u < \frac{4 π}{3} + 2 πn

\frac{2 π}{3} + 2 πn < u < \frac{4 π}{3} + 2 πn

Substitute back

2 x = u

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

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

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

a < u < b

then

a < u and u < b

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

\frac{2 π}{3} + 2 πn < 2 x : x > \frac{π}{3} + πn

\frac{2 π}{3} + 2 πn < 2 x

Switch sides

2 x > \frac{2 π}{3} + 2 πn

Divide both sides by

2

2 x > \frac{2 π}{3} + 2 πn

Divide both sides by

2

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

Simplify

\frac{2 x}{2} > \frac{\frac{2 π}{3}}{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{2 π}{3}}{2} + \frac{2 πn}{2} : \frac{π}{3} + πn

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

\frac{\frac{2 π}{3}}{2} = \frac{π}{3}

\frac{\frac{2 π}{3}}{2}

Apply the fraction rule:

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

= \frac{2 π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{2 π}{6}

Cancel the common factor:

2

= \frac{π}{3}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{3} + πn

x > \frac{π}{3} + πn

x > \frac{π}{3} + πn

x > \frac{π}{3} + πn

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

2 x < \frac{4 π}{3} + 2 πn

Divide both sides by

2

2 x < \frac{4 π}{3} + 2 πn

Divide both sides by

2

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

Simplify

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

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

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

\frac{\frac{4 π}{3}}{2}

Apply the fraction rule:

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

= \frac{4 π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{4 π}{6}

Cancel the common factor:

2

= \frac{2 π}{3}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{2 π}{3} + πn

x < \frac{2 π}{3} + πn

x < \frac{2 π}{3} + πn

x < \frac{2 π}{3} + πn

Combine the intervals

x > \frac{π}{3} + πn and x < \frac{2 π}{3} + πn

Merge Overlapping Intervals

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

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

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