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(cos^2(x)-1/2)/(tan(x)-sqrt(3))<0

Solution

\frac{cos ^{2} ( x ) - \frac{1}{2}}{tan ( x ) - 3} < 0

Solution

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

Interval Notation

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

Decimal Notation

πn \leq x < 0.78539 \dots + πn or 1.04719 \dots + πn < x < 1.57079 \dots + πn or 2.35619 \dots + πn < x \leq 3.14159 \dots + πn

Solution steps

\frac{cos ^{2} ( x ) - \frac{1}{2}}{tan ( x ) - 3} < 0

Use the following identity:

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

Therefore

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

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

Simplify

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

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

Multiply by the conjugate

\frac{tan ( x ) + 3}{tan ( x ) + 3}

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

Simplify

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

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

Join

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

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

Combine the fractions

1 - \frac{1}{2} : \frac{1}{2}

1 - \frac{1}{2}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

1 \cdot 2 - 1 = 1

1 \cdot 2 - 1

Multiply the numbers:

1 \cdot 2 = 2

= 2 - 1

Subtract the numbers:

2 - 1 = 1

= 1

= \frac{1}{2}

= - sin^{2} (x)

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

(tan (x) - 3) (tan (x) + 3) = tan^{2} (x) - 3

(tan (x) - 3) (tan (x) + 3)

Apply Difference of Two Squares Formula:

(a - b) (a + b) = a^{2} - b^{2}

a = tan (x), b = 3

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

(3)^{2} = 3

(3)^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

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

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 3

= tan^{2} (x) - 3

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

Factor

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

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

Factor out common term

- 1

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

Factor

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

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

Apply radical rule:

a = (a)^{2}

\frac{1}{2} = (\frac{1}{2})^{2}

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

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

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

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

Factor

tan^{2} (x) - 3 : (tan (x) + 3) (tan (x) - 3)

tan^{2} (x) - 3

Apply radical rule:

a = (a)^{2}

3 = (3)^{2}

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

tan^{2} (x) - (3)^{2} = (tan (x) + 3) (tan (x) - 3)

= (tan (x) + 3) (tan (x) - 3)

= - \frac{( sin ( x ) + \frac{1}{2} ) ( sin ( x ) - \frac{1}{2} ) ( tan ( x ) + 3 )}{( tan ( x ) + 3 ) ( tan ( x ) - 3 )}

Cancel the common factor:

tan (x) + 3

= - \frac{( sin ( x ) + \frac{1}{2} ) ( sin ( x ) - \frac{1}{2} )}{tan ( x ) - 3}

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

Periodicity of

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

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

is composed of the following functions and periods:

sin (x)

with periodicity of

2 π

The compound periodicity is:

= π

Express with sin, cos

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

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

- \frac{( sin ( x ) + \frac{1}{2} ) ( sin ( x ) - \frac{1}{2} )}{\frac{s i n ( x )}{c o s ( x )} - 3} < 0

- \frac{( sin ( x ) + \frac{1}{2} ) ( sin ( x ) - \frac{1}{2} )}{\frac{s i n ( x )}{c o s ( x )} - 3} < 0

Simplify

- \frac{( sin ( x ) + \frac{1}{2} ) ( sin ( x ) - \frac{1}{2} )}{\frac{s i n ( x )}{c o s ( x )} - 3} : - \frac{cos ( x ) ( sin ( x ) + \frac{1}{2} ) ( sin ( x ) - \frac{1}{2} )}{sin ( x ) - 3 cos ( x )}

- \frac{( sin ( x ) + \frac{1}{2} ) ( sin ( x ) - \frac{1}{2} )}{\frac{s i n ( x )}{c o s ( x )} - 3}

Join

\frac{sin ( x )}{cos ( x )} - 3 : \frac{sin ( x ) - 3 cos ( x )}{cos ( x )}

\frac{sin ( x )}{cos ( x )} - 3

Convert element to fraction:

3 = \frac{3 cos ( x )}{cos ( x )}

= \frac{sin ( x )}{cos ( x )} - \frac{3 cos ( x )}{cos ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{sin ( x ) - 3 cos ( x )}{cos ( x )}

= - \frac{( sin ( x ) + \frac{1}{2} ) ( sin ( x ) - \frac{1}{2} )}{\frac{s i n ( x ) - 3 c o s ( x )}{c o s ( x )}}

Apply the fraction rule:

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

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

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

Find the zeroes and undifined points of

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

for

0 \leq x < π

To find the zeroes, set the inequality to zero

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

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

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

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

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

Solving each part separately

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

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

cos (x) = 0, 0 \leq x < π

General solutions for

cos (x) = 0

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}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

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

Solutions for the range

0 \leq x < π

x = \frac{π}{2}

sin (x) + \frac{1}{2} = 0, 0 \leq x < π :

No Solution

sin (x) + \frac{1}{2} = 0, 0 \leq x < π

Move

\frac{1}{2}

to the right side

sin (x) + \frac{1}{2} = 0

Subtract

\frac{1}{2}

from both sides

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

Simplify

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

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

Apply trig inverse properties

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

General solutions for

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

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{1}{2}) + 2 πn, x = π + arcsin (\frac{1}{2}) + 2 πn

x = arcsin (- \frac{1}{2}) + 2 πn, x = π + arcsin (\frac{1}{2}) + 2 πn

Solutions for the range

0 \leq x < π

No Solution

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

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

Move

\frac{1}{2}

to the right side

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

Add

\frac{1}{2}

to both sides

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

Simplify

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

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

Apply trig inverse properties

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

General solutions for

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

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{1}{2}) + 2 πn, x = π - arcsin (\frac{1}{2}) + 2 πn

x = arcsin (\frac{1}{2}) + 2 πn, x = π - arcsin (\frac{1}{2}) + 2 πn

Solutions for the range

0 \leq x < π

x = \frac{π}{4}, x = \frac{3 π}{4}

Combine all the solutions

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

Find the undefined points:

x = \frac{π}{3}

Find the zeros of the denominator

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

Rewrite using trig identities

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

Divide both sides by

cos (x), cos (x) \neq = 0

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

Simplify

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

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

tan (x) - 3 = 0

tan (x) - 3 = 0

Move

3

to the right side

tan (x) - 3 = 0

Add

3

to both sides

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

Simplify

tan (x) = 3

tan (x) = 3

General solutions for

tan (x) = 3

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{3} + πn

x = \frac{π}{3} + πn

Solutions for the range

0 \leq x < π

x = \frac{π}{3}

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

Identify the intervals

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

Summarize in a table:

cos (x) sin (x) + \frac{1}{2} sin (x) - \frac{1}{2} sin (x) - 3 cos (x) - \frac{c o s ( x ) ( s i n ( x ) + \frac{1}{2} ) ( s i n ( x ) - \frac{1}{2} )}{s i n ( x ) - 3 c o s ( x )} x = 0 + + - - - 0 < x < \frac{π}{4} + + - - - x = \frac{π}{4} + + 0 - 0 \frac{π}{4} < x < \frac{π}{3} + + + - + x = \frac{π}{3} + + + 0 Undefined \frac{π}{3} < x < \frac{π}{2} + + + + - x = \frac{π}{2} 0 + + + 0 \frac{π}{2} < x < \frac{3 π}{4} - + + + + x = \frac{3 π}{4} - + 0 + 0 \frac{3 π}{4} < x < π - + - + - x = π - + - + -

Identify the intervals that satisfy the required condition:

< 0

x = 0 or 0 < x < \frac{π}{4} or \frac{π}{3} < x < \frac{π}{2} or \frac{3 π}{4} < x < π or x = π

Merge Overlapping Intervals

0 \leq x < \frac{π}{4} or \frac{π}{3} < x < \frac{π}{2} or \frac{3 π}{4} < x < π or x = π

The union of two intervals is the set of numbers which are in either interval

x = 0

0 < x < \frac{π}{4}

0 \leq x < \frac{π}{4}

The union of two intervals is the set of numbers which are in either interval

0 \leq x < \frac{π}{4}

\frac{π}{3} < x < \frac{π}{2}

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

The union of two intervals is the set of numbers which are in either interval

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

\frac{3 π}{4} < x < π

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

The union of two intervals is the set of numbers which are in either interval

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

x = π

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

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

Apply the periodicity of

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

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

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