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(2sin(2x)-1)/(cos(2x)-3cos(x)+2)>= 0

Solution

\frac{2 sin ( 2 x ) - 1}{cos ( 2 x ) - 3 cos ( x ) + 2} \geq 0

Solution

2 πn < x \leq \frac{π}{12} + 2 πn or \frac{π}{3} + 2 πn < x \leq \frac{5 π}{12} + 2 πn or \frac{13 π}{12} + 2 πn \leq x \leq \frac{17 π}{12} + 2 πn or \frac{5 π}{3} + 2 πn < x < 2 π + 2 πn

Interval Notation

(2 πn, \frac{π}{12} + 2 πn] \cup (\frac{π}{3} + 2 πn, \frac{5 π}{12} + 2 πn] \cup [\frac{13 π}{12} + 2 πn, \frac{17 π}{12} + 2 πn] \cup (\frac{5 π}{3} + 2 πn, 2 π + 2 πn)

Decimal Notation

2 πn < x \leq 0.26179 \dots + 2 πn or 1.04719 \dots + 2 πn < x \leq 1.30899 \dots + 2 πn or 3.40339 \dots + 2 πn \leq x \leq 4.45058 \dots + 2 πn or 5.23598 \dots + 2 πn < x < 6.28318 \dots + 2 πn

Solution steps

\frac{2 sin ( 2 x ) - 1}{cos ( 2 x ) - 3 cos ( x ) + 2} \geq 0

Use the following identity:

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

\frac{- 1 + 2 sin ( 2 x )}{2 + cos ^{2} ( x ) - sin ^{2} ( x ) - 3 cos ( x )} \geq 0

Periodicity of

\frac{- 1 + 2 sin ( 2 x )}{2 + cos ^{2} ( x ) - sin ^{2} ( x ) - 3 cos ( x )} : 2 π

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

is composed of the following functions and periods:

sin (2 x)

with periodicity of

\frac{2 π}{2}

The compound periodicity is:

= 2 π

Find the zeroes and undifined points of

\frac{- 1 + 2 sin ( 2 x )}{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{- 1 + 2 sin ( 2 x )}{2 + cos ^{2} ( x ) - sin ^{2} ( x ) - 3 cos ( x )} = 0

\frac{- 1 + 2 sin ( 2 x )}{2 + cos ^{2} ( x ) - sin ^{2} ( x ) - 3 cos ( x )} = 0, 0 \leq x < 2 π : x = \frac{π}{12}, x = \frac{5 π}{12}, x = \frac{13 π}{12}, x = \frac{17 π}{12}

\frac{- 1 + 2 sin ( 2 x )}{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

- 1 + 2 sin (2 x) = 0

Move

1

to the right side

- 1 + 2 sin (2 x) = 0

Add

1

to both sides

- 1 + 2 sin (2 x) + 1 = 0 + 1

Simplify

2 sin (2 x) = 1

2 sin (2 x) = 1

Divide both sides by

2

2 sin (2 x) = 1

Divide both sides by

2

\frac{2 sin ( 2 x )}{2} = \frac{1}{2}

Simplify

sin (2 x) = \frac{1}{2}

sin (2 x) = \frac{1}{2}

General solutions for

sin (2 x) = \frac{1}{2}

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

2 x = \frac{π}{6} + 2 πn, 2 x = \frac{5 π}{6} + 2 πn

2 x = \frac{π}{6} + 2 πn, 2 x = \frac{5 π}{6} + 2 πn

Solve

2 x = \frac{π}{6} + 2 πn : x = \frac{π}{12} + πn

2 x = \frac{π}{6} + 2 πn

Divide both sides by

2

2 x = \frac{π}{6} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{π}{6}}{2} + \frac{2 πn}{2} : \frac{π}{12} + πn

\frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{6}}{2} = \frac{π}{12}

\frac{\frac{π}{6}}{2}

Apply the fraction rule:

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

= \frac{π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{12} + πn

x = \frac{π}{12} + πn

x = \frac{π}{12} + πn

x = \frac{π}{12} + πn

Solve

2 x = \frac{5 π}{6} + 2 πn : x = \frac{5 π}{12} + πn

2 x = \frac{5 π}{6} + 2 πn

Divide both sides by

2

2 x = \frac{5 π}{6} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2} : \frac{5 π}{12} + πn

\frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{6}}{2} = \frac{5 π}{12}

\frac{\frac{5 π}{6}}{2}

Apply the fraction rule:

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

= \frac{5 π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{5 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

Solutions for the range

0 \leq x < 2 π

x = \frac{π}{12}, x = \frac{5 π}{12}, x = \frac{13 π}{12}, x = \frac{17 π}{12}

Find the undefined points:

x = 0, x = \frac{π}{3}, x = \frac{5 π}{3}

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

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

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 3)^{2} = 3^{2}

= 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} : 1

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

Apply rule

- (- a) = a

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

Add the numbers:

3 + 1 = 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

3 - 1 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

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

cos (x) = 1, 0 \leq x < 2 π : x = 0

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 = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

Solutions for the range

0 \leq x < 2 π

x = 0

cos (x) = \frac{1}{2}, 0 \leq x < 2 π : x = \frac{π}{3}, x = \frac{5 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{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Solutions for the range

0 \leq x < 2 π

x = \frac{π}{3}, x = \frac{5 π}{3}

Combine all the solutions

x = 0, x = \frac{π}{3}, x = \frac{5 π}{3}

0, \frac{π}{12}, \frac{π}{3}, \frac{5 π}{12}, \frac{13 π}{12}, \frac{17 π}{12}, \frac{5 π}{3}

Identify the intervals

0 < x < \frac{π}{12}, \frac{π}{12} < x < \frac{π}{3}, \frac{π}{3} < x < \frac{5 π}{12}, \frac{5 π}{12} < x < \frac{13 π}{12}, \frac{13 π}{12} < x < \frac{17 π}{12}, \frac{17 π}{12} < x < \frac{5 π}{3}, \frac{5 π}{3} < x < 2 π

Summarize in a table:

- 1 + 2 sin (2 x) 2 + cos^{2} (x) - sin^{2} (x) - 3 cos (x) \frac{- 1 + 2 s i n ( 2 x )}{2 + c o s ^{2} ( x ) - s i n ^{2} ( x ) - 3 c o s ( x )} x = 0 - 0 Undefined 0 < x < \frac{π}{12} - - + x = \frac{π}{12} 0 - 0 \frac{π}{12} < x < \frac{π}{3} + - - x = \frac{π}{3} + 0 Undefined \frac{π}{3} < x < \frac{5 π}{12} + + + x = \frac{5 π}{12} 0 + 0 \frac{5 π}{12} < x < \frac{13 π}{12} - + - x = \frac{13 π}{12} 0 + 0 \frac{13 π}{12} < x < \frac{17 π}{12} + + + x = \frac{17 π}{12} 0 + 0 \frac{17 π}{12} < x < \frac{5 π}{3} - + - x = \frac{5 π}{3} - 0 Undefined \frac{5 π}{3} < x < 2 π - - + x = 2 π - 0 Undefined

Identify the intervals that satisfy the required condition:

\geq 0

0 < x < \frac{π}{12} or x = \frac{π}{12} or \frac{π}{3} < x < \frac{5 π}{12} or x = \frac{5 π}{12} or x = \frac{13 π}{12} or \frac{13 π}{12} < x < \frac{17 π}{12} or x = \frac{17 π}{12} or \frac{5 π}{3} < x < 2 π

Merge Overlapping Intervals

0 < x \leq \frac{π}{12} or \frac{π}{3} < x \leq \frac{5 π}{12} or \frac{13 π}{12} \leq x \leq \frac{17 π}{12} or \frac{5 π}{3} < x < 2 π