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Popular Trigonometry >

cos(x)+cos(2x)>0

Solution

cos (x) + cos (2 x) > 0

Solution

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

Interval Notation

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

Decimal

- 1.04719 \dots + 2 πn < x < 1.04719 \dots + 2 πn

Solution steps

cos (x) + cos (2 x) > 0

Use the following identity:

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

- 1 + 2 cos^{2} (x) + cos (x) > 0

Let:

u = cos (x)

- 1 + 2 u^{2} + u > 0

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

- 1 + 2 u^{2} + u > 0

Factor

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

- 1 + 2 u^{2} + u

Write in the standard form

a x^{2} + b x + c

= 2 u^{2} + u - 1

Break the expression into groups

2 u^{2} + u - 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

FalseCheck

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

True

u = 2, v = - 1

Group into

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

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

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

Factor out

u

from

2 u^{2} - u : u (2 u - 1)

2 u^{2} - u

Apply exponent rule:

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

u^{2} = uu

= 2 uu - u

Factor out common term

u

= u (2 u - 1)

= u (2 u - 1) + (2 u - 1)

Factor out common term

2 u - 1

= (2 u - 1) (u + 1)

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

Identify the intervals

Find the signs of the factors of

(2 u - 1) (u + 1)

Find the signs of

2 u - 1

2 u - 1 = 0 : u = \frac{1}{2}

2 u - 1 = 0

Move

1

to the right side

2 u - 1 = 0

Add

1

to both sides

2 u - 1 + 1 = 0 + 1

Simplify

2 u = 1

2 u = 1

Divide both sides by

2

2 u = 1

Divide both sides by

2

\frac{2 u}{2} = \frac{1}{2}

Simplify

u = \frac{1}{2}

u = \frac{1}{2}

2 u - 1 < 0 : u < \frac{1}{2}

2 u - 1 < 0

Move

1

to the right side

2 u - 1 < 0

Add

1

to both sides

2 u - 1 + 1 < 0 + 1

Simplify

2 u < 1

2 u < 1

Divide both sides by

2

2 u < 1

Divide both sides by

2

\frac{2 u}{2} < \frac{1}{2}

Simplify

u < \frac{1}{2}

u < \frac{1}{2}

2 u - 1 > 0 : u > \frac{1}{2}

2 u - 1 > 0

Move

1

to the right side

2 u - 1 > 0

Add

1

to both sides

2 u - 1 + 1 > 0 + 1

Simplify

2 u > 1

2 u > 1

Divide both sides by

2

2 u > 1

Divide both sides by

2

\frac{2 u}{2} > \frac{1}{2}

Simplify

u > \frac{1}{2}

u > \frac{1}{2}

Find the signs of

u + 1

u + 1 = 0 : u = - 1

u + 1 = 0

Move

1

to the right side

u + 1 = 0

Subtract

1

from both sides

u + 1 - 1 = 0 - 1

Simplify

u = - 1

u = - 1

u + 1 < 0 : u < - 1

u + 1 < 0

Move

1

to the right side

u + 1 < 0

Subtract

1

from both sides

u + 1 - 1 < 0 - 1

Simplify

u < - 1

u < - 1

u + 1 > 0 : u > - 1

u + 1 > 0

Move

1

to the right side

u + 1 > 0

Subtract

1

from both sides

u + 1 - 1 > 0 - 1

Simplify

u > - 1

u > - 1

Summarize in a table:

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

Identify the intervals that satisfy the required condition:

> 0

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

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

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

Substitute back

u = cos (x)

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

cos (x) < - 1 :

False for all

x \in R

cos (x) < - 1

Range of

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

Function range definition

The range of the basic

cos

function is

- 1 \leq cos (x) \leq 1

- 1 \leq cos (x) \leq 1

cos (x) < - 1 and - 1 \leq cos (x) \leq 1 :

False

Let

y = cos (x)

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 x \in R

False for all x \in R

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

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

For

cos (x) > a

, if

- 1 \leq a < 1

then

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

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

Simplify

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

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

Use the following trivial identity:

arccos (\frac{1}{2}) = \frac{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{π}{3}

Simplify

arccos (\frac{1}{2}) : \frac{π}{3}

arccos (\frac{1}{2})

Use the following trivial identity:

arccos (\frac{1}{2}) = \frac{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{π}{3}

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

Combine the intervals

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

Merge Overlapping Intervals

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

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