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Popular >

(2cos(x)-sqrt(3))/(cos^2(x))<0

Solution

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

Solution

\frac{π}{6} + 2 πn < x < \frac{π}{2} + 2 πn or \frac{π}{2} + 2 πn < x < \frac{3 π}{2} + 2 πn or \frac{3 π}{2} + 2 πn < x < \frac{11 π}{6} + 2 πn

Interval Notation

(\frac{π}{6} + 2 πn, \frac{π}{2} + 2 πn) \cup (\frac{π}{2} + 2 πn, \frac{3 π}{2} + 2 πn) \cup (\frac{3 π}{2} + 2 πn, \frac{11 π}{6} + 2 πn)

Decimal

0.52359 \dots + 2 πn < x < 1.57079 \dots + 2 πn or 1.57079 \dots + 2 πn < x < 4.71238 \dots + 2 πn or 4.71238 \dots + 2 πn < x < 5.75958 \dots + 2 πn

Solution steps

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

Let:

u = cos (x)

\frac{2 u - 3}{u ^{2}} < 0

\frac{2 u - 3}{u ^{2}} < 0 : u < 0 or 0 < u < \frac{3}{2}

\frac{2 u - 3}{u ^{2}} < 0

Identify the intervals

Find the signs of the factors of

\frac{2 u - 3}{u ^{2}}

Find the signs of

2 u - 3

2 u - 3 = 0 : u = \frac{3}{2}

2 u - 3 = 0

Move

3

to the right side

2 u - 3 = 0

Add

3

to both sides

2 u - 3 + 3 = 0 + 3

Simplify

2 u = 3

2 u = 3

Divide both sides by

2

2 u = 3

Divide both sides by

2

\frac{2 u}{2} = \frac{3}{2}

Simplify

u = \frac{3}{2}

u = \frac{3}{2}

2 u - 3 < 0 : u < \frac{3}{2}

2 u - 3 < 0

Move

3

to the right side

2 u - 3 < 0

Add

3

to both sides

2 u - 3 + 3 < 0 + 3

Simplify

2 u < 3

2 u < 3

Divide both sides by

2

2 u < 3

Divide both sides by

2

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

Simplify

u < \frac{3}{2}

u < \frac{3}{2}

2 u - 3 > 0 : u > \frac{3}{2}

2 u - 3 > 0

Move

3

to the right side

2 u - 3 > 0

Add

3

to both sides

2 u - 3 + 3 > 0 + 3

Simplify

2 u > 3

2 u > 3

Divide both sides by

2

2 u > 3

Divide both sides by

2

\frac{2 u}{2} > \frac{3}{2}

Simplify

u > \frac{3}{2}

u > \frac{3}{2}

Find the signs of

u^{2}

u^{2} = 0 : u = 0

u^{2} = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

u = 0

u^{2} > 0 : u < 0 or u > 0

u^{2} > 0

For

u^{n} > 0

, if

n

is even then

u < 0

u > 0

u < 0 or u > 0

Find singularity points

Find the zeros of the denominator

u^{2} : u = 0

u^{2} = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

u = 0

Summarize in a table:

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

Identify the intervals that satisfy the required condition:

< 0

u < 0 or 0 < u < \frac{3}{2}

u < 0 or 0 < u < \frac{3}{2}

u < 0 or 0 < u < \frac{3}{2}

Substitute back

u = cos (x)

cos (x) < 0 or 0 < cos (x) < \frac{3}{2}

cos (x) < 0 : \frac{π}{2} + 2 πn < x < \frac{3 π}{2} + 2 πn

cos (x) < 0

For

cos (x) < a

, if

- 1 < a \leq 1

then

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

arccos (0) + 2 πn < x < 2 π - arccos (0) + 2 πn

Simplify

arccos (0) : \frac{π}{2}

arccos (0)

Use the following trivial identity:

arccos (0) = \frac{π}{2}

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}{2}

Simplify

2 π - arccos (0) : \frac{3 π}{2}

2 π - arccos (0)

Use the following trivial identity:

arccos (0) = \frac{π}{2}

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{π}{2}

Simplify

2 π - \frac{π}{2}

Convert element to fraction:

2 π = \frac{2 π 2}{2}

= \frac{2 π 2}{2} - \frac{π}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{2 π 2 - π}{2}

2 π 2 - π = 3 π

2 π 2 - π

Multiply the numbers:

2 \cdot 2 = 4

= 4 π - π

Add similar elements:

4 π - π = 3 π

= 3 π

= \frac{3 π}{2}

= \frac{3 π}{2}

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

0 < cos (x) < \frac{3}{2} : \frac{π}{6} + 2 πn < x < \frac{π}{2} + 2 πn or \frac{3 π}{2} + 2 πn < x < \frac{11 π}{6} + 2 πn

0 < cos (x) < \frac{3}{2}

a < u < b

then

a < u and u < b

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

0 < cos (x) : - \frac{π}{2} + 2 πn < x < \frac{π}{2} + 2 πn

0 < cos (x)

Switch sides

cos (x) > 0

For

cos (x) > a

, if

- 1 \leq a < 1

then

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

- arccos (0) + 2 πn < x < arccos (0) + 2 πn

Simplify

- arccos (0) : - \frac{π}{2}

- arccos (0)

Use the following trivial identity:

arccos (0) = \frac{π}{2}

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}{2}

Simplify

arccos (0) : \frac{π}{2}

arccos (0)

Use the following trivial identity:

arccos (0) = \frac{π}{2}

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}{2}

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

cos (x) < \frac{3}{2} : \frac{π}{6} + 2 πn < x < \frac{11 π}{6} + 2 πn

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

For

cos (x) < a

, if

- 1 < a \leq 1

then

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

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

Simplify

arccos (\frac{3}{2}) : \frac{π}{6}

arccos (\frac{3}{2})

Use the following trivial identity:

arccos (\frac{3}{2}) = \frac{π}{6}

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}{6}

Simplify

2 π - arccos (\frac{3}{2}) : \frac{11 π}{6}

2 π - arccos (\frac{3}{2})

Use the following trivial identity:

arccos (\frac{3}{2}) = \frac{π}{6}

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{π}{6}

Simplify

2 π - \frac{π}{6}

Convert element to fraction:

2 π = \frac{2 π 6}{6}

= \frac{2 π 6}{6} - \frac{π}{6}

Since the denominators are equal, combine the fractions:

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

= \frac{2 π 6 - π}{6}

2 π 6 - π = 11 π

2 π 6 - π

Multiply the numbers:

2 \cdot 6 = 12

= 12 π - π

Add similar elements:

12 π - π = 11 π

= 11 π

= \frac{11 π}{6}

= \frac{11 π}{6}

\frac{π}{6} + 2 πn < x < \frac{11 π}{6} + 2 πn

Combine the intervals

- \frac{π}{2} + 2 πn < x < \frac{π}{2} + 2 πn and \frac{π}{6} + 2 πn < x < \frac{11 π}{6} + 2 πn

Merge Overlapping Intervals

\frac{π}{6} + 2 πn < x < \frac{π}{2} + 2 πn or \frac{3 π}{2} + 2 πn < x < \frac{11 π}{6} + 2 πn

Combine the intervals

\frac{π}{2} + 2 πn < x < \frac{3 π}{2} + 2 πn or (\frac{π}{6} + 2 πn < x < \frac{π}{2} + 2 πn or \frac{3 π}{2} + 2 πn < x < \frac{11 π}{6} + 2 πn)

Merge Overlapping Intervals

\frac{π}{6} + 2 πn < x < \frac{π}{2} + 2 πn or \frac{π}{2} + 2 πn < x < \frac{3 π}{2} + 2 πn or \frac{3 π}{2} + 2 πn < x < \frac{11 π}{6} + 2 πn

Popular Examples

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