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Popular Trigonometry >

(4cos(x)+3)/(3cos(x)+1)<2

Solution

\frac{4 cos ( x ) + 3}{3 cos ( x ) + 1} < 2

Solution

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

Interval Notation

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

Decimal

- 1.04719 \dots + 2 πn < x < 1.04719 \dots + 2 πn or 1.91063 \dots + 2 πn < x < 4.37255 \dots + 2 πn

Solution steps

\frac{4 cos ( x ) + 3}{3 cos ( x ) + 1} < 2

Let:

u = cos (x)

\frac{4 u + 3}{3 u + 1} < 2

\frac{4 u + 3}{3 u + 1} < 2 : u < - \frac{1}{3} or u > \frac{1}{2}

\frac{4 u + 3}{3 u + 1} < 2

Rewrite in standard form

\frac{4 u + 3}{3 u + 1} < 2

Subtract

2

from both sides

\frac{4 u + 3}{3 u + 1} - 2 < 2 - 2

Simplify

\frac{4 u + 3}{3 u + 1} - 2 < 0

Simplify

\frac{4 u + 3}{3 u + 1} - 2 : \frac{- 2 u + 1}{3 u + 1}

\frac{4 u + 3}{3 u + 1} - 2

Convert element to fraction:

2 = \frac{2 ( 3 u + 1 )}{3 u + 1}

= \frac{4 u + 3}{3 u + 1} - \frac{2 ( 3 u + 1 )}{3 u + 1}

Since the denominators are equal, combine the fractions:

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

= \frac{4 u + 3 - 2 ( 3 u + 1 )}{3 u + 1}

Expand

4 u + 3 - 2 (3 u + 1) : - 2 u + 1

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

Expand

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

- 2 (3 u + 1)

Apply the distributive law:

a (b + c) = ab + a c

a = - 2, b = 3 u, c = 1

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

Apply minus-plus rules

+ (- a) = - a

= - 2 \cdot 3 u - 2 \cdot 1

Simplify

- 2 \cdot 3 u - 2 \cdot 1 : - 6 u - 2

- 2 \cdot 3 u - 2 \cdot 1

Multiply the numbers:

2 \cdot 3 = 6

= - 6 u - 2 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= - 6 u - 2

= - 6 u - 2

= 4 u + 3 - 6 u - 2

Simplify

4 u + 3 - 6 u - 2 : - 2 u + 1

4 u + 3 - 6 u - 2

Group like terms

= 4 u - 6 u + 3 - 2

Add similar elements:

4 u - 6 u = - 2 u

= - 2 u + 3 - 2

Add/Subtract the numbers:

3 - 2 = 1

= - 2 u + 1

= - 2 u + 1

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

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

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

Identify the intervals

Find the signs of the factors of

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

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

Multiply both sides by

- 1

- 2 u < - 1

Multiply both sides by -1 (reverse the inequality)

(- 2 u) (- 1) > (- 1) (- 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

Multiply both sides by

- 1

- 2 u > - 1

Multiply both sides by -1 (reverse the inequality)

(- 2 u) (- 1) < (- 1) (- 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

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

Subtract

1

from 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

Subtract

1

from 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

Subtract

1

from 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 singularity points

Find the zeros of the denominator

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

3 u + 1 = 0

Move

1

to the right side

3 u + 1 = 0

Subtract

1

from 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}

Summarize in a table:

- 2 u + 1 3 u + 1 \frac{- 2 u + 1}{3 u + 1} u < - \frac{1}{3} + - - u = - \frac{1}{3} + 0 Undefined - \frac{1}{3} < u < \frac{1}{2} + + + u = \frac{1}{2} 0 + 0 u > \frac{1}{2} - + -

Identify the intervals that satisfy the required condition:

< 0

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

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

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

Substitute back

u = cos (x)

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

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

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

cos (x) > \frac{1}{2}

For

cos (x) > a

, if

- 1 \leq a < 1

then

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

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

Simplify

- arccos (\frac{1}{2}) : - \frac{π}{3}

- arccos (\frac{1}{2})

Use the following trivial identity:

arccos (\frac{1}{2}) = \frac{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{π}{3}

Simplify

arccos (\frac{1}{2}) : \frac{π}{3}

arccos (\frac{1}{2})

Use the following trivial identity:

arccos (\frac{1}{2}) = \frac{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{π}{3}

- \frac{π}{3} + 2 πn < x < \frac{π}{3} + 2 πn

Combine the intervals

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

Merge Overlapping Intervals

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

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