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2cos^2(x)-3cos(x)-2>= 0

Solution

2 cos^{2} (x) - 3 cos (x) - 2 \geq 0

Solution

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

Interval Notation

[\frac{2 π}{3} + 2 πn, \frac{4 π}{3} + 2 πn]

Decimal Notation

2.09439 \dots + 2 πn \leq x \leq 4.18879 \dots + 2 πn

Solution steps

2 cos^{2} (x) - 3 cos (x) - 2 \geq 0

Let:

u = cos (x)

2 u^{2} - 3 u - 2 \geq 0

2 u^{2} - 3 u - 2 \geq 0 : u \leq - \frac{1}{2} or u \geq 2

2 u^{2} - 3 u - 2 \geq 0

Factor

2 u^{2} - 3 u - 2 : (2 u + 1) (u - 2)

2 u^{2} - 3 u - 2

Break the expression into groups

2 u^{2} - 3 u - 2

Definition

Factors of

4 : 1, 2, 4

4

Divisors (Factors)

Find the Prime factors of

4 : 2, 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2

Add the prime factors:

2

Add 1 and the number

4

itself

1, 4

The factors of

4

1, 2, 4

Negative factors of

4 : - 1, - 2, - 4

Multiply the factors by

- 1

to get the negative factors

- 1, - 2, - 4

For every two factors such that

u * v = - 4,

check if

u + v = - 3

Check

u = 1, v = - 4 : u * v = - 4, u + v = - 3 \Rightarrow

TrueCheck

u = 2, v = - 2 : u * v = - 4, u + v = 0 \Rightarrow

False

u = 1, v = - 4

Group into

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

(2 u^{2} + u) + (- 4 u - 2)

= (2 u^{2} + u) + (- 4 u - 2)

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)

Factor out

- 2

from

- 4 u - 2 : - 2 (2 u + 1)

- 4 u - 2

Rewrite

4

2 \cdot 2

= - 2 \cdot 2 u - 2

Factor out common term

- 2

= - 2 (2 u + 1)

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

Factor out common term

2 u + 1

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

(2 u + 1) (u - 2) \geq 0

Identify the intervals

Find the signs of the factors of

(2 u + 1) (u - 2)

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

Subtract

1

from 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

Subtract

1

from 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

Subtract

1

from 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 - 2

u - 2 = 0 : u = 2

u - 2 = 0

Move

2

to the right side

u - 2 = 0

Add

2

to 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

Add

2

to 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

Add

2

to both sides

u - 2 + 2 > 0 + 2

Simplify

u > 2

u > 2

Summarize in a table:

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

Identify the intervals that satisfy the required condition:

\geq 0

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

Merge Overlapping Intervals

u \leq - \frac{1}{2} or u = 2 or u > 2

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

u < - \frac{1}{2}

u = - \frac{1}{2}

u \leq - \frac{1}{2}

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

u \leq - \frac{1}{2}

u = 2

u \leq - \frac{1}{2} or u = 2

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

u \leq - \frac{1}{2} or u = 2

u > 2

u \leq - \frac{1}{2} or u \geq 2

u \leq - \frac{1}{2} or u \geq 2

u \leq - \frac{1}{2} or u \geq 2

u \leq - \frac{1}{2} or u \geq 2

Substitute back

u = cos (x)

cos (x) \leq - \frac{1}{2} or cos (x) \geq 2

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

cos (x) \leq - \frac{1}{2}

For

cos (x) \leq a

, if

- 1 < a < 1

then

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

arccos (- \frac{1}{2}) + 2 πn \leq x \leq 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 \leq x \leq \frac{4 π}{3} + 2 πn

cos (x) \geq 2 :

False for all

x \in R

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 :

False

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

False for all y \in R

False for all y \in R

No Solution for x \in R

False for all x \in R

Combine the intervals

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

Merge Overlapping Intervals

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

Popular Examples

pi/2 cos(pi/2 x)>= 0

\frac{π}{2} cos (\frac{π}{2} x) \geq 0

cos((x-45))< 1/2 ,0<= x<= 360

cos ((x - 45)^{\circ}) < \frac{1}{2}, 0^{\circ} \leq x \leq 36 0^{\circ}

cos(x)<= (sqrt(2))/2

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

cos(φ)<= (sqrt(2))/2

cos (φ) \leq \frac{2}{2}

2cos(y)>0

2 cos (y) > 0