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(cos(3x)-sqrt(3))/(cos(2x)+3cos(x)+2)>0

Solution

\frac{cos ( 3 x ) - 3}{cos ( 2 x ) + 3 cos ( x ) + 2} > 0

Solution

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

Interval Notation

(\frac{2 π}{3} + 2 πn, π + 2 πn) \cup (π + 2 πn, \frac{4 π}{3} + 2 πn)

Decimal Notation

2.09439 \dots + 2 πn < x < 3.14159 \dots + 2 πn or 3.14159 \dots + 2 πn < x < 4.18879 \dots + 2 πn

Solution steps

\frac{cos ( 3 x ) - 3}{cos ( 2 x ) + 3 cos ( x ) + 2} > 0

Use the following identity:

cos (2 x) = cos^{2} (x) - sin^{2} (x)

\frac{cos ( 3 x ) - 3}{2 + cos ^{2} ( x ) - sin ^{2} ( x ) + 3 cos ( x )} > 0

Periodicity of

\frac{cos ( 3 x ) - 3}{2 + cos ^{2} ( x ) - sin ^{2} ( x ) + 3 cos ( x )} : 2 π

\frac{cos ( 3 x ) - 3}{2 + cos ^{2} ( x ) - sin ^{2} ( x ) + 3 cos ( x )}

is composed of the following functions and periods:

cos (3 x)

with periodicity of

\frac{2 π}{3}

The compound periodicity is:

= 2 π

Find the zeroes and undifined points of

\frac{cos ( 3 x ) - 3}{2 + cos ^{2} ( x ) - sin ^{2} ( x ) + 3 cos ( x )}

for

0 \leq x < 2 π

To find the zeroes, set the inequality to zero

\frac{cos ( 3 x ) - 3}{2 + cos ^{2} ( x ) - sin ^{2} ( x ) + 3 cos ( x )} = 0

\frac{cos ( 3 x ) - 3}{2 + cos ^{2} ( x ) - sin ^{2} ( x ) + 3 cos ( x )} = 0, 0 \leq x < 2 π :

No Solution for

x \in R

\frac{cos ( 3 x ) - 3}{2 + cos ^{2} ( x ) - sin ^{2} ( x ) + 3 cos ( x )} = 0, 0 \leq x < 2 π

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

cos (3 x) - 3 = 0

Move

3

to the right side

cos (3 x) - 3 = 0

Add

3

to both sides

cos (3 x) - 3 + 3 = 0 + 3

Simplify

cos (3 x) = 3

cos (3 x) = 3

- 1 \leq cos (x) \leq 1

No Solution for x \in R

Find the undefined points:

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

Find the zeros of the denominator

2 + cos^{2} (x) - sin^{2} (x) + 3 cos (x) = 0

Rewrite using trig identities

2 + cos^{2} (x) - sin^{2} (x) + 3 cos (x)

Use the Pythagorean identity:

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

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

= 2 + cos^{2} (x) - (1 - cos^{2} (x)) + 3 cos (x)

Simplify

2 + cos^{2} (x) - (1 - cos^{2} (x)) + 3 cos (x) : 2 cos^{2} (x) + 3 cos (x) + 1

2 + cos^{2} (x) - (1 - cos^{2} (x)) + 3 cos (x)

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

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

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

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

= 2 + cos^{2} (x) - 1 + cos^{2} (x) + 3 cos (x)

Simplify

2 + cos^{2} (x) - 1 + cos^{2} (x) + 3 cos (x) : 2 cos^{2} (x) + 3 cos (x) + 1

2 + cos^{2} (x) - 1 + cos^{2} (x) + 3 cos (x)

Group like terms

= cos^{2} (x) + cos^{2} (x) + 3 cos (x) + 2 - 1

Add similar elements:

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

= 2 cos^{2} (x) + 3 cos (x) + 2 - 1

Add/Subtract the numbers:

2 - 1 = 1

= 2 cos^{2} (x) + 3 cos (x) + 1

= 2 cos^{2} (x) + 3 cos (x) + 1

= 2 cos^{2} (x) + 3 cos (x) + 1

1 + 2 cos^{2} (x) + 3 cos (x) = 0

Solve by substitution

1 + 2 cos^{2} (x) + 3 cos (x) = 0

Let:

cos (x) = u

1 + 2 u^{2} + 3 u = 0

1 + 2 u^{2} + 3 u = 0 : u = - \frac{1}{2}, u = - 1

1 + 2 u^{2} + 3 u = 0

Write in the standard form

a x^{2} + b x + c = 0

2 u^{2} + 3 u + 1 = 0

Solve with the quadratic formula

2 u^{2} + 3 u + 1 = 0

Quadratic Equation Formula:

For

a = 2, b = 3, c = 1

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

3^{2} - 4 \cdot 2 \cdot 1 = 1

3^{2} - 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- 3 \pm 1}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- 3 + 1}{2 \cdot 2}, u_{2} = \frac{- 3 - 1}{2 \cdot 2}

u = \frac{- 3 + 1}{2 \cdot 2} : - \frac{1}{2}

\frac{- 3 + 1}{2 \cdot 2}

Add/Subtract the numbers:

- 3 + 1 = - 2

= \frac{- 2}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2}{4}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2}{4}

Cancel the common factor:

2

= - \frac{1}{2}

u = \frac{- 3 - 1}{2 \cdot 2} : - 1

\frac{- 3 - 1}{2 \cdot 2}

Subtract the numbers:

- 3 - 1 = - 4

= \frac{- 4}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{4}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

cos (x) = - \frac{1}{2}, cos (x) = - 1

cos (x) = - \frac{1}{2}, cos (x) = - 1

cos (x) = - \frac{1}{2}, 0 \leq x < 2 π : x = \frac{2 π}{3}, x = \frac{4 π}{3}

cos (x) = - \frac{1}{2}, 0 \leq x < 2 π

General solutions for

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

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Solutions for the range

0 \leq x < 2 π

x = \frac{2 π}{3}, x = \frac{4 π}{3}

cos (x) = - 1, 0 \leq x < 2 π : x = π

cos (x) = - 1, 0 \leq x < 2 π

General solutions for

cos (x) = - 1

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = π + 2 πn

x = π + 2 πn

Solutions for the range

0 \leq x < 2 π

x = π

Combine all the solutions

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

\frac{2 π}{3}, π, \frac{4 π}{3}

Identify the intervals

0 < x < \frac{2 π}{3}, \frac{2 π}{3} < x < π, π < x < \frac{4 π}{3}, \frac{4 π}{3} < x < 2 π

Summarize in a table:

cos (3 x) - 3 2 + cos^{2} (x) - sin^{2} (x) + 3 cos (x) \frac{c o s ( 3 x ) - 3}{2 + c o s ^{2} ( x ) - s i n ^{2} ( x ) + 3 c o s ( x )} x = 0 - + - 0 < x < \frac{2 π}{3} - + - x = \frac{2 π}{3} - 0 Undefined \frac{2 π}{3} < x < π - - + x = π - 0 Undefined π < x < \frac{4 π}{3} - - + x = \frac{4 π}{3} - 0 Undefined \frac{4 π}{3} < x < 2 π - + - x = 2 π - + -

Identify the intervals that satisfy the required condition:

> 0

\frac{2 π}{3} < x < π or π < x < \frac{4 π}{3}

Apply the periodicity of

\frac{cos ( 3 x ) - 3}{2 + cos ^{2} ( x ) - sin ^{2} ( x ) + 3 cos ( x )}

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

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