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Popular Trigonometry >

cos(x)(cos(x)+2)<= 0

Solution

cos (x) (cos (x) + 2) \leq 0

Solution

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

Interval Notation

[\frac{π}{2} + 2 πn, \frac{3 π}{2} + 2 πn]

Decimal

1.57079 \dots + 2 πn \leq x \leq 4.71238 \dots + 2 πn

Solution steps

cos (x) (cos (x) + 2) \leq 0

Let:

u = cos (x)

u (u + 2) \leq 0

u (u + 2) \leq 0 : - 2 \leq u \leq 0

u (u + 2) \leq 0

Identify the intervals

Find the signs of the factors of

u (u + 2)

Find the signs of

u

u = 0

u < 0

u > 0

Find the signs of

u + 2

u + 2 = 0 : u = - 2

u + 2 = 0

Move

2

to the right side

u + 2 = 0

Subtract

2

from both sides

u + 2 - 2 = 0 - 2

Simplify

u = - 2

u = - 2

u + 2 < 0 : u < - 2

u + 2 < 0

Move

2

to the right side

u + 2 < 0

Subtract

2

from both sides

u + 2 - 2 < 0 - 2

Simplify

u < - 2

u < - 2

u + 2 > 0 : u > - 2

u + 2 > 0

Move

2

to the right side

u + 2 > 0

Subtract

2

from both sides

u + 2 - 2 > 0 - 2

Simplify

u > - 2

u > - 2

Summarize in a table:

u u + 2 u (u + 2) u < - 2 - - + u = - 2 - 00 - 2 < u < 0 - + - u = 0 0 + 0 u > 0 + + +

Identify the intervals that satisfy the required condition:

\leq 0

u = - 2 or - 2 < u < 0 or u = 0

Merge Overlapping Intervals

- 2 \leq u < 0 or u = 0

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

u = - 2

- 2 < u < 0

- 2 \leq u < 0

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

- 2 \leq u < 0

u = 0

- 2 \leq u \leq 0

- 2 \leq u \leq 0

- 2 \leq u \leq 0

- 2 \leq u \leq 0

Substitute back

u = cos (x)

- 2 \leq cos (x) \leq 0

a \leq u \leq b

then

a \leq u and u \leq b

- 2 \leq cos (x) and cos (x) \leq 0

- 2 \leq cos (x) :

True for all

x \in R

- 2 \leq cos (x)

Switch sides

cos (x) \geq - 2

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) \geq - 2 and - 1 \leq cos (x) \leq 1 : - 1 \leq cos (x) \leq 1

Let

y = cos (x)

Combine the intervals

y \geq - 2 and - 1 \leq y \leq 1

Merge Overlapping Intervals

y \geq - 2 and - 1 \leq y \leq 1

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

y \geq - 2

and

- 1 \leq y \leq 1

- 1 \leq y \leq 1

- 1 \leq y \leq 1

True for all x

True for all x \in R

cos (x) \leq 0 : \frac{π}{2} + 2 πn \leq x \leq \frac{3 π}{2} + 2 πn

cos (x) \leq 0

For

cos (x) \leq a

, if

- 1 < a < 1

then

arccos (a) + 2 πn \leq x \leq 2 π - arccos (a) + 2 πn

arccos (0) + 2 πn \leq x \leq 2 π - arccos (0) + 2 πn

Simplify

arccos (0) : \frac{π}{2}

arccos (0)

Use the following trivial identity:

arccos (0) = \frac{π}{2}

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}{2}

Simplify

2 π - arccos (0) : \frac{3 π}{2}

2 π - arccos (0)

Use the following trivial identity:

arccos (0) = \frac{π}{2}

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{π}{2}

Simplify

2 π - \frac{π}{2}

Convert element to fraction:

2 π = \frac{2 π 2}{2}

= \frac{2 π 2}{2} - \frac{π}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{2 π 2 - π}{2}

2 π 2 - π = 3 π

2 π 2 - π

Multiply the numbers:

2 \cdot 2 = 4

= 4 π - π

Add similar elements:

4 π - π = 3 π

= 3 π

= \frac{3 π}{2}

= \frac{3 π}{2}

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

Combine the intervals

True for all x \in R and \frac{π}{2} + 2 πn \leq x \leq \frac{3 π}{2} + 2 πn

Merge Overlapping Intervals

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

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