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Popular >

((4*cos^2(x)-3))/((sin(x)+cos(x)+5))<0

Solution

\frac{( 4 \cdot cos ^{2} ( x ) - 3 )}{( sin ( x ) + cos ( x ) + 5 )} < 0

Solution

\frac{π}{6} + 2 πn < x < \frac{5 π}{6} + 2 πn or \frac{7 π}{6} + 2 πn < x < \frac{11 π}{6} + 2 πn

Interval Notation

(\frac{π}{6} + 2 πn, \frac{5 π}{6} + 2 πn) \cup (\frac{7 π}{6} + 2 πn, \frac{11 π}{6} + 2 πn)

Decimal Notation

0.52359 \dots + 2 πn < x < 2.61799 \dots + 2 πn or 3.66519 \dots + 2 πn < x < 5.75958 \dots + 2 πn

Solution steps

\frac{4 cos ^{2} ( x ) - 3}{sin ( x ) + cos ( x ) + 5} < 0

Use the following identity:

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

Therefore

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

\frac{4 ( 1 - sin ^{2} ( x ) ) - 3}{sin ( x ) + cos ( x ) + 5} < 0

Simplify

\frac{4 ( 1 - sin ^{2} ( x ) ) - 3}{sin ( x ) + cos ( x ) + 5} : \frac{- 4 sin ^{2} ( x ) + 1}{sin ( x ) + cos ( x ) + 5}

\frac{4 ( 1 - sin ^{2} ( x ) ) - 3}{sin ( x ) + cos ( x ) + 5}

Expand

4 (1 - sin^{2} (x)) - 3 : - 4 sin^{2} (x) + 1

4 (1 - sin^{2} (x)) - 3

Expand

4 (1 - sin^{2} (x)) : 4 - 4 sin^{2} (x)

4 (1 - sin^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 1, c = sin^{2} (x)

= 4 \cdot 1 - 4 sin^{2} (x)

Multiply the numbers:

4 \cdot 1 = 4

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

= 4 - 4 sin^{2} (x) - 3

Simplify

4 - 4 sin^{2} (x) - 3 : - 4 sin^{2} (x) + 1

4 - 4 sin^{2} (x) - 3

Group like terms

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

Add/Subtract the numbers:

4 - 3 = 1

= - 4 sin^{2} (x) + 1

= - 4 sin^{2} (x) + 1

= \frac{- 4 sin ^{2} ( x ) + 1}{sin ( x ) + cos ( x ) + 5}

\frac{- 4 sin ^{2} ( x ) + 1}{sin ( x ) + cos ( x ) + 5} < 0

Periodicity of

\frac{- 4 sin ^{2} ( x ) + 1}{sin ( x ) + cos ( x ) + 5} : 2 π

\frac{- 4 sin ^{2} ( x ) + 1}{sin ( x ) + cos ( x ) + 5}

is composed of the following functions and periods:

sin (x)

with periodicity of

2 π

The compound periodicity is:

= 2 π

Find the zeroes and undifined points of

\frac{- 4 sin ^{2} ( x ) + 1}{sin ( x ) + cos ( x ) + 5}

for

0 \leq x < 2 π

To find the zeroes, set the inequality to zero

\frac{- 4 sin ^{2} ( x ) + 1}{sin ( x ) + cos ( x ) + 5} = 0

\frac{- 4 sin ^{2} ( x ) + 1}{sin ( x ) + cos ( x ) + 5} = 0, 0 \leq x < 2 π : x = \frac{π}{6}, x = \frac{5 π}{6}, x = \frac{7 π}{6}, x = \frac{11 π}{6}

\frac{- 4 sin ^{2} ( x ) + 1}{sin ( x ) + cos ( x ) + 5} = 0, 0 \leq x < 2 π

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

- 4 sin^{2} (x) + 1 = 0

Solve by substitution

- 4 sin^{2} (x) + 1 = 0

Let:

sin (x) = u

- 4 u^{2} + 1 = 0

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

- 4 u^{2} + 1 = 0

Move

1

to the right side

- 4 u^{2} + 1 = 0

Subtract

1

from both sides

- 4 u^{2} + 1 - 1 = 0 - 1

Simplify

- 4 u^{2} = - 1

- 4 u^{2} = - 1

Divide both sides by

- 4

- 4 u^{2} = - 1

Divide both sides by

- 4

\frac{- 4 u ^{2}}{- 4} = \frac{- 1}{- 4}

Simplify

u^{2} = \frac{1}{4}

u^{2} = \frac{1}{4}

For

x^{2} = f (a)

the solutions are

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

u = \frac{1}{4}, u = - \frac{1}{4}

\frac{1}{4} = \frac{1}{2}

\frac{1}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{1}{2}

Apply rule

1 = 1

= \frac{1}{2}

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

- \frac{1}{4}

Simplify

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

\frac{1}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{1}{2}

= - \frac{1}{2}

Apply rule

1 = 1

= - \frac{1}{2}

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}{6}, x = \frac{5 π}{6}

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}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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

Solutions for the range

0 \leq x < 2 π

x = \frac{π}{6}, x = \frac{5 π}{6}

sin (x) = - \frac{1}{2}, 0 \leq x < 2 π : x = \frac{7 π}{6}, x = \frac{11 π}{6}

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{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Solutions for the range

0 \leq x < 2 π

x = \frac{7 π}{6}, x = \frac{11 π}{6}

Combine all the solutions

x = \frac{π}{6}, x = \frac{5 π}{6}, x = \frac{7 π}{6}, x = \frac{11 π}{6}

Find the undefined points:

No Solution

Find the zeros of the denominator

sin (x) + cos (x) + 5 = 0

Rewrite using trig identities

sin (x) + cos (x) + 5

sin (x) + cos (x) = 2 sin (x + \frac{π}{4})

sin (x) + cos (x)

Rewrite as

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

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{1}{2}

Use the following trivial identity:

sin (\frac{π}{4}) = \frac{1}{2}

= 2 (cos (\frac{π}{4}) sin (x) + sin (\frac{π}{4}) cos (x))

Use the Angle Sum identity:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= 2 sin (x + \frac{π}{4})

= 5 + 2 sin (x + \frac{π}{4})

5 + 2 sin (x + \frac{π}{4}) = 0

Move

5

to the right side

5 + 2 sin (x + \frac{π}{4}) = 0

Subtract

5

from both sides

5 + 2 sin (x + \frac{π}{4}) - 5 = 0 - 5

Simplify

2 sin (x + \frac{π}{4}) = - 5

2 sin (x + \frac{π}{4}) = - 5

Divide both sides by

2

2 sin (x + \frac{π}{4}) = - 5

Divide both sides by

2

\frac{2 sin ( x + \frac{π}{4} )}{2} = \frac{- 5}{2}

Simplify

\frac{2 sin ( x + \frac{π}{4} )}{2} = \frac{- 5}{2}

Simplify

\frac{2 sin ( x + \frac{π}{4} )}{2} : sin (x + \frac{π}{4})

\frac{2 sin ( x + \frac{π}{4} )}{2}

Cancel the common factor:

2

= sin (x + \frac{π}{4})

Simplify

\frac{- 5}{2} : - \frac{5 2}{2}

\frac{- 5}{2}

Apply the fraction rule:

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

= - \frac{5}{2}

Rationalize

- \frac{5}{2} : - \frac{5 2}{2}

- \frac{5}{2}

Multiply by the conjugate

\frac{2}{2}

= - \frac{5 2}{2 2}

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= - \frac{5 2}{2}

= - \frac{5 2}{2}

sin (x + \frac{π}{4}) = - \frac{5 2}{2}

sin (x + \frac{π}{4}) = - \frac{5 2}{2}

sin (x + \frac{π}{4}) = - \frac{5 2}{2}

- 1 \leq sin (x) \leq 1

No Solution for x \in R

\frac{π}{6}, \frac{5 π}{6}, \frac{7 π}{6}, \frac{11 π}{6}

Identify the intervals

0 < x < \frac{π}{6}, \frac{π}{6} < x < \frac{5 π}{6}, \frac{5 π}{6} < x < \frac{7 π}{6}, \frac{7 π}{6} < x < \frac{11 π}{6}, \frac{11 π}{6} < x < 2 π

Summarize in a table:

- 4 sin^{2} (x) + 1 sin (x) + cos (x) + 5 \frac{- 4 s i n ^{2} ( x ) + 1}{s i n ( x ) + c o s ( x ) + 5} x = 0 + + + 0 < x < \frac{π}{6} + + + x = \frac{π}{6} 0 + 0 \frac{π}{6} < x < \frac{5 π}{6} - + - x = \frac{5 π}{6} 0 + 0 \frac{5 π}{6} < x < \frac{7 π}{6} + + + x = \frac{7 π}{6} 0 + 0 \frac{7 π}{6} < x < \frac{11 π}{6} - + - x = \frac{11 π}{6} 0 + 0 \frac{11 π}{6} < x < 2 π + + + x = 2 π + + +

Identify the intervals that satisfy the required condition:

< 0

\frac{π}{6} < x < \frac{5 π}{6} or \frac{7 π}{6} < x < \frac{11 π}{6}

Apply the periodicity of

\frac{- 4 sin ^{2} ( x ) + 1}{sin ( x ) + cos ( x ) + 5}

\frac{π}{6} + 2 πn < x < \frac{5 π}{6} + 2 πn or \frac{7 π}{6} + 2 πn < x < \frac{11 π}{6} + 2 πn

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