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Popular Trigonometry >

3cos^2(x)+5cos(x)-2<0

Solution

3 cos^{2} (x) + 5 cos (x) - 2 < 0

Solution

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

Interval Notation

(arccos (\frac{1}{3}) + 2 πn, 2 π - arccos (\frac{1}{3}) + 2 πn)

Decimal

1.23095 \dots + 2 πn < x < 5.05222 \dots + 2 πn

Solution steps

3 cos^{2} (x) + 5 cos (x) - 2 < 0

Let:

u = cos (x)

3 u^{2} + 5 u - 2 < 0

3 u^{2} + 5 u - 2 < 0 : - 2 < u < \frac{1}{3}

3 u^{2} + 5 u - 2 < 0

Factor

3 u^{2} + 5 u - 2 : (3 u - 1) (u + 2)

3 u^{2} + 5 u - 2

Break the expression into groups

3 u^{2} + 5 u - 2

Definition

Factors of

6 : 1, 2, 3, 6

6

Divisors (Factors)

Find the Prime factors of

6 : 2, 3

6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 3

Add the prime factors:

2, 3

Add 1 and the number

6

itself

1, 6

The factors of

6

1, 2, 3, 6

Negative factors of

6 : - 1, - 2, - 3, - 6

Multiply the factors by

- 1

to get the negative factors

- 1, - 2, - 3, - 6

For every two factors such that

u * v = - 6,

check if

u + v = 5

Check

u = 1, v = - 6 : u * v = - 6, u + v = - 5 \Rightarrow

FalseCheck

u = 2, v = - 3 : u * v = - 6, u + v = - 1 \Rightarrow

False

u = 6, v = - 1

Group into

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

(3 u^{2} - u) + (6 u - 2)

= (3 u^{2} - u) + (6 u - 2)

Factor out

u

from

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

3 u^{2} - u

Apply exponent rule:

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

u^{2} = uu

= 3 uu - u

Factor out common term

u

= u (3 u - 1)

Factor out

2

from

6 u - 2 : 2 (3 u - 1)

6 u - 2

Rewrite

6

2 \cdot 3

= 2 \cdot 3 u - 2

Factor out common term

2

= 2 (3 u - 1)

= u (3 u - 1) + 2 (3 u - 1)

Factor out common term

3 u - 1

= (3 u - 1) (u + 2)

(3 u - 1) (u + 2) < 0

Identify the intervals

Find the signs of the factors of

(3 u - 1) (u + 2)

Find the signs of

3 u - 1

3 u - 1 = 0 : u = \frac{1}{3}

3 u - 1 = 0

Move

1

to the right side

3 u - 1 = 0

Add

1

to both sides

3 u - 1 + 1 = 0 + 1

Simplify

3 u = 1

3 u = 1

Divide both sides by

3

3 u = 1

Divide both sides by

3

\frac{3 u}{3} = \frac{1}{3}

Simplify

u = \frac{1}{3}

u = \frac{1}{3}

3 u - 1 < 0 : u < \frac{1}{3}

3 u - 1 < 0

Move

1

to the right side

3 u - 1 < 0

Add

1

to both sides

3 u - 1 + 1 < 0 + 1

Simplify

3 u < 1

3 u < 1

Divide both sides by

3

3 u < 1

Divide both sides by

3

\frac{3 u}{3} < \frac{1}{3}

Simplify

u < \frac{1}{3}

u < \frac{1}{3}

3 u - 1 > 0 : u > \frac{1}{3}

3 u - 1 > 0

Move

1

to the right side

3 u - 1 > 0

Add

1

to both sides

3 u - 1 + 1 > 0 + 1

Simplify

3 u > 1

3 u > 1

Divide both sides by

3

3 u > 1

Divide both sides by

3

\frac{3 u}{3} > \frac{1}{3}

Simplify

u > \frac{1}{3}

u > \frac{1}{3}

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:

3 u - 1 u + 2 (3 u - 1) (u + 2) u < - 2 - - + u = - 2 - 00 - 2 < u < \frac{1}{3} - + - u = \frac{1}{3} 0 + 0 u > \frac{1}{3} + + +

Identify the intervals that satisfy the required condition:

< 0

- 2 < u < \frac{1}{3}

- 2 < u < \frac{1}{3}

- 2 < u < \frac{1}{3}

Substitute back

u = cos (x)

- 2 < cos (x) < \frac{1}{3}

a < u < b

then

a < u and u < b

- 2 < cos (x) and cos (x) < \frac{1}{3}

- 2 < cos (x) :

True for all

x \in R

- 2 < cos (x)

Switch sides

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

Let

y = cos (x)

Combine the intervals

y > - 2 and - 1 \leq y \leq 1

Merge Overlapping Intervals

y > - 2 and - 1 \leq y \leq 1

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

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

cos (x) < \frac{1}{3}

For

cos (x) < a

, if

- 1 < a \leq 1

then

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

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

Combine the intervals

True for all x \in R and arccos (\frac{1}{3}) + 2 πn < x < 2 π - arccos (\frac{1}{3}) + 2 πn

Merge Overlapping Intervals

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

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