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

Solution

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

Solution

2 πn \leq x < \frac{π}{4} + 2 πn or \frac{π}{2} + 2 πn \leq x < \frac{3 π}{4} + 2 πn or \frac{3 π}{2} + 2 πn \leq x < \frac{7 π}{4} + 2 πn or \frac{7 π}{4} + 2 πn < x \leq 2 π + 2 πn

Interval Notation

[2 πn, \frac{π}{4} + 2 πn) \cup [\frac{π}{2} + 2 πn, \frac{3 π}{4} + 2 πn) \cup [\frac{3 π}{2} + 2 πn, \frac{7 π}{4} + 2 πn) \cup (\frac{7 π}{4} + 2 πn, 2 π + 2 πn]

Decimal Notation

2 πn \leq x < 0.78539 \dots + 2 πn or 1.57079 \dots + 2 πn \leq x < 2.35619 \dots + 2 πn or 4.71238 \dots + 2 πn \leq x < 5.49778 \dots + 2 πn or 5.49778 \dots + 2 πn < x \leq 6.28318 \dots + 2 πn

Solution steps

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

Simplify

\frac{3 cos ( x )}{2 - 2 sin ( x )} : \frac{3 cos ( x ) ( 2 + 2 sin ( x ) )}{2 - 4 sin ^{2} ( x )}

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

Multiply by the conjugate

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

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

(2 - 2 sin (x)) (2 + 2 sin (x)) = 2 - 4 sin^{2} (x)

(2 - 2 sin (x)) (2 + 2 sin (x))

Apply Difference of Two Squares Formula:

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

a = 2, b = 2 sin (x)

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

Simplify

(2)^{2} - (2 sin (x))^{2} : 2 - 4 sin^{2} (x)

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

(2)^{2} = 2

(2)^{2}

Apply radical rule:

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

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

Apply exponent rule:

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

= 2^{\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

= 2

(2 sin (x))^{2} = 4 sin^{2} (x)

(2 sin (x))^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

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

2^{2} = 4

= 4 sin^{2} (x)

= 2 - 4 sin^{2} (x)

= 2 - 4 sin^{2} (x)

= \frac{3 cos ( x ) ( 2 + 2 sin ( x ) )}{2 - 4 sin ^{2} ( x )}

\frac{3 cos ( x ) ( 2 + 2 sin ( x ) )}{2 - 4 sin ^{2} ( x )} \geq 0

Periodicity of

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

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

is composed of the following functions and periods:

cos (x)

with periodicity of

2 π

The compound periodicity is:

= 2 π

Find the zeroes and undifined points of

\frac{3 cos ( x ) ( 2 + 2 sin ( x ) )}{2 - 4 sin ^{2} ( x )}

for

0 \leq x < 2 π

To find the zeroes, set the inequality to zero

\frac{3 cos ( x ) ( 2 + 2 sin ( x ) )}{2 - 4 sin ^{2} ( x )} = 0

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

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

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

3 cos (x) (2 + 2 sin (x)) = 0

Solving each part separately

cos (x) = 0 or 2 + 2 sin (x) = 0

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

cos (x) = 0, 0 \leq x < 2 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 < 2 π

x = \frac{π}{2}, x = \frac{3 π}{2}

2 + 2 sin (x) = 0, 0 \leq x < 2 π : x = \frac{5 π}{4}, x = \frac{7 π}{4}

2 + 2 sin (x) = 0, 0 \leq x < 2 π

Move

2

to the right side

2 + 2 sin (x) = 0

Subtract

2

from both sides

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

Simplify

2 sin (x) = - 2

2 sin (x) = - 2

Divide both sides by

2

2 sin (x) = - 2

Divide both sides by

2

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

Simplify

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

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

General solutions for

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

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Solutions for the range

0 \leq x < 2 π

x = \frac{5 π}{4}, x = \frac{7 π}{4}

Combine all the solutions

x = \frac{π}{2}, x = \frac{3 π}{2}, x = \frac{5 π}{4}, x = \frac{7 π}{4}

Since the equation is undefined for:

\frac{5 π}{4}, \frac{7 π}{4}

x = \frac{π}{2}, x = \frac{3 π}{2}

Find the undefined points:

x = \frac{π}{4}, x = \frac{3 π}{4}, x = \frac{5 π}{4}, x = \frac{7 π}{4}

Find the zeros of the denominator

2 - 4 sin^{2} (x) = 0

Solve by substitution

2 - 4 sin^{2} (x) = 0

Let:

sin (x) = u

2 - 4 u^{2} = 0

2 - 4 u^{2} = 0 : u = \frac{1}{2}, u = - \frac{1}{2}

2 - 4 u^{2} = 0

Move

2

to the right side

2 - 4 u^{2} = 0

Subtract

2

from both sides

2 - 4 u^{2} - 2 = 0 - 2

Simplify

- 4 u^{2} = - 2

- 4 u^{2} = - 2

Divide both sides by

- 4

- 4 u^{2} = - 2

Divide both sides by

- 4

\frac{- 4 u ^{2}}{- 4} = \frac{- 2}{- 4}

Simplify

\frac{- 4 u ^{2}}{- 4} = \frac{- 2}{- 4}

Simplify

\frac{- 4 u ^{2}}{- 4} : u^{2}

\frac{- 4 u ^{2}}{- 4}

Apply the fraction rule:

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

= \frac{4 u ^{2}}{4}

Divide the numbers:

\frac{4}{4} = 1

= u^{2}

Simplify

\frac{- 2}{- 4} : \frac{1}{2}

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

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

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

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

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

Substitute back

u = sin (x)

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

sin (x) = \frac{1}{2}, 0 \leq x < 2 π

General solutions for

sin (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}

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

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

Solutions for the range

0 \leq x < 2 π

x = \frac{π}{4}, x = \frac{3 π}{4}

sin (x) = - \frac{1}{2}, 0 \leq x < 2 π : x = \frac{5 π}{4}, x = \frac{7 π}{4}

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

General solutions for

sin (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}

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Solutions for the range

0 \leq x < 2 π

x = \frac{5 π}{4}, x = \frac{7 π}{4}

Combine all the solutions

x = \frac{π}{4}, x = \frac{3 π}{4}, x = \frac{5 π}{4}, x = \frac{7 π}{4}

\frac{π}{4}, \frac{π}{2}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{3 π}{2}, \frac{7 π}{4}

Identify the intervals

0 < x < \frac{π}{4}, \frac{π}{4} < x < \frac{π}{2}, \frac{π}{2} < x < \frac{3 π}{4}, \frac{3 π}{4} < x < \frac{5 π}{4}, \frac{5 π}{4} < x < \frac{3 π}{2}, \frac{3 π}{2} < x < \frac{7 π}{4}, \frac{7 π}{4} < x < 2 π

Summarize in a table:

cos (x) 2 + 2 sin (x) 2 - 4 sin^{2} (x) \frac{3 c o s ( x ) ( 2 + 2 s i n ( x ) )}{2 - 4 s i n ^{2} ( x )} x = 0 + + + + 0 < x < \frac{π}{4} + + + + x = \frac{π}{4} + + 0 Undefined \frac{π}{4} < x < \frac{π}{2} + + - - x = \frac{π}{2} 0 + - 0 \frac{π}{2} < x < \frac{3 π}{4} - + - + x = \frac{3 π}{4} - + 0 Undefined \frac{3 π}{4} < x < \frac{5 π}{4} - + + - x = \frac{5 π}{4} - 00 Undefined \frac{5 π}{4} < x < \frac{3 π}{2} - - - - x = \frac{3 π}{2} 0 - - 0 \frac{3 π}{2} < x < \frac{7 π}{4} + - - + x = \frac{7 π}{4} + 00 Undefined \frac{7 π}{4} < x < 2 π + + + + x = 2 π + + + +

Identify the intervals that satisfy the required condition:

\geq 0

x = 0 or 0 < x < \frac{π}{4} or x = \frac{π}{2} or \frac{π}{2} < x < \frac{3 π}{4} or x = \frac{3 π}{2} or \frac{3 π}{2} < x < \frac{7 π}{4} or \frac{7 π}{4} < x < 2 π or x = 2 π

Merge Overlapping Intervals

0 \leq x < \frac{π}{4} or \frac{π}{2} \leq x < \frac{3 π}{4} or \frac{3 π}{2} \leq x < \frac{7 π}{4} or \frac{7 π}{4} < x < 2 π or x = 2 π