English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

6cos^2(x)>tan(x)cos(x)+4

Solution

6 cos^{2} (x) > tan (x) cos (x) + 4

Solution

2 πn \leq x < 1.16841 \dots + 2 πn or π - 1.16841 \dots + 2 πn < x < \frac{3 π}{2} + 2 πn or \frac{3 π}{2} + 2 πn < x \leq 2 π + 2 πn

Interval Notation

Decimal

2 πn \leq x < 1.16841 \dots + 2 πn or 1.97317 \dots + 2 πn < x < 4.71238 \dots + 2 πn or 4.71238 \dots + 2 πn < x \leq 6.28318 \dots + 2 πn

Solution steps

6 cos^{2} (x) > tan (x) cos (x) + 4

Move

tan (x) cos (x)

to the left side

6 cos^{2} (x) > tan (x) cos (x) + 4

Subtract

tan (x) cos (x)

from both sides

6 cos^{2} (x) - tan (x) cos (x) > tan (x) cos (x) + 4 - tan (x) cos (x)

6 cos^{2} (x) - tan (x) cos (x) > 4

6 cos^{2} (x) - tan (x) cos (x) > 4

Factor

6 cos^{2} (x) - tan (x) cos (x) : cos (x) (6 cos (x) - tan (x))

6 cos^{2} (x) - tan (x) cos (x)

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

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

= 6 cos (x) cos (x) - cos (x) tan (x)

Factor out common term

cos (x)

= cos (x) (6 cos (x) - tan (x))

cos (x) (6 cos (x) - tan (x)) > 4

To find the zeroes, set the inequality to zero

cos (x) (6 cos (x) - tan (x)) = 0

Solve

cos (x) (6 cos (x) - tan (x)) = 0

for

0 \leq x < 2 π

cos (x) (6 cos (x) - tan (x)) = 0

Solving each part separately

cos (x) = 0 : x = \frac{π}{2} or 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}

6 cos (x) - tan (x) = 0 : x = 1.16841 \dots or x = π - 1.16841 \dots

6 cos (x) - tan (x) = 0, 0 \leq x < 2 π

Express with sin, cos

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

Simplify

6 cos (x) - \frac{sin ( x )}{cos ( x )} : \frac{6 cos ^{2} ( x ) - sin ( x )}{cos ( x )}

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

Convert element to fraction:

6 cos (x) = \frac{6 cos ( x ) cos ( x )}{cos ( x )}

= \frac{6 cos ( x ) cos ( x )}{cos ( x )} - \frac{sin ( x )}{cos ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{6 cos ( x ) cos ( x ) - sin ( x )}{cos ( x )}

6 cos (x) cos (x) - sin (x) = 6 cos^{2} (x) - sin (x)

6 cos (x) cos (x) - sin (x)

6 cos (x) cos (x) = 6 cos^{2} (x)

6 cos (x) cos (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= 6 cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= 6 cos^{2} (x)

= 6 cos^{2} (x) - sin (x)

= \frac{6 cos ^{2} ( x ) - sin ( x )}{cos ( x )}

\frac{6 cos ^{2} ( x ) - sin ( x )}{cos ( x )} = 0

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

6 cos^{2} (x) - sin (x) = 0

Add

sin (x)

to both sides

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

Square both sides

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

Subtract

sin^{2} (x)

from both sides

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

Factor

36 cos^{4} (x) - sin^{2} (x) : (6 cos^{2} (x) + sin (x)) (6 cos^{2} (x) - sin (x))

36 cos^{4} (x) - sin^{2} (x)

Rewrite

36 cos^{4} (x) - sin^{2} (x)

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

36 cos^{4} (x) - sin^{2} (x)

Rewrite

36

6^{2}

= 6^{2} cos^{4} (x) - sin^{2} (x)

Apply exponent rule:

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

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

= 6^{2} (cos^{2} (x))^{2} - sin^{2} (x)

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

6^{2} (cos^{2} (x))^{2} = (6 cos^{2} (x))^{2}

= (6 cos^{2} (x))^{2} - sin^{2} (x)

= (6 cos^{2} (x))^{2} - sin^{2} (x)

Apply Difference of Two Squares Formula:

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

(6 cos^{2} (x))^{2} - sin^{2} (x) = (6 cos^{2} (x) + sin (x)) (6 cos^{2} (x) - sin (x))

= (6 cos^{2} (x) + sin (x)) (6 cos^{2} (x) - sin (x))

(6 cos^{2} (x) + sin (x)) (6 cos^{2} (x) - sin (x)) = 0

Solving each part separately

6 cos^{2} (x) + sin (x) = 0 or 6 cos^{2} (x) - sin (x) = 0

6 cos^{2} (x) + sin (x) = 0, 0 \leq x < 2 π : x = π + arcsin (\frac{145 - 1}{12}), x = - arcsin (\frac{145 - 1}{12}) + 2 π

6 cos^{2} (x) + sin (x) = 0, 0 \leq x < 2 π

Rewrite using trig identities

sin (x) + 6 cos^{2} (x)

Use the Pythagorean identity:

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

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

= sin (x) + 6 (1 - sin^{2} (x))

sin (x) + (1 - sin^{2} (x)) \cdot 6 = 0

Solve by substitution

sin (x) + (1 - sin^{2} (x)) \cdot 6 = 0

Let:

sin (x) = u

u + (1 - u^{2}) \cdot 6 = 0

u + (1 - u^{2}) \cdot 6 = 0 : u = - \frac{- 1 + 145}{12}, u = \frac{1 + 145}{12}

u + (1 - u^{2}) \cdot 6 = 0

Expand

u + (1 - u^{2}) \cdot 6 : u + 6 - 6 u^{2}

u + (1 - u^{2}) \cdot 6

= u + 6 (1 - u^{2})

Expand

6 (1 - u^{2}) : 6 - 6 u^{2}

6 (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = 6, b = 1, c = u^{2}

= 6 \cdot 1 - 6 u^{2}

Multiply the numbers:

6 \cdot 1 = 6

= 6 - 6 u^{2}

= u + 6 - 6 u^{2}

u + 6 - 6 u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

- 6 u^{2} + u + 6 = 0

Solve with the quadratic formula

- 6 u^{2} + u + 6 = 0

Quadratic Equation Formula:

For

a = - 6, b = 1, c = 6

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 6 ) \cdot 6}{2 ( - 6 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 6 ) \cdot 6}{2 ( - 6 )}

1^{2} - 4 (- 6) \cdot 6 = 145

1^{2} - 4 (- 6) \cdot 6

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 6) \cdot 6

Apply rule

- (- a) = a

= 1 + 4 \cdot 6 \cdot 6

Multiply the numbers:

4 \cdot 6 \cdot 6 = 144

= 1 + 144

Add the numbers:

1 + 144 = 145

= 145

u_{1, 2} = \frac{- 1 \pm 145}{2 ( - 6 )}

Separate the solutions

u_{1} = \frac{- 1 + 145}{2 ( - 6 )}, u_{2} = \frac{- 1 - 145}{2 ( - 6 )}

u = \frac{- 1 + 145}{2 ( - 6 )} : - \frac{- 1 + 145}{12}

\frac{- 1 + 145}{2 ( - 6 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 145}{- 2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 1 + 145}{- 12}

Apply the fraction rule:

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

= - \frac{- 1 + 145}{12}

u = \frac{- 1 - 145}{2 ( - 6 )} : \frac{1 + 145}{12}

\frac{- 1 - 145}{2 ( - 6 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 145}{- 2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 1 - 145}{- 12}

Apply the fraction rule:

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

- 1 - 145 = - (1 + 145)

= \frac{1 + 145}{12}

The solutions to the quadratic equation are:

u = - \frac{- 1 + 145}{12}, u = \frac{1 + 145}{12}

Substitute back

u = sin (x)

sin (x) = - \frac{- 1 + 145}{12}, sin (x) = \frac{1 + 145}{12}

sin (x) = - \frac{- 1 + 145}{12}, sin (x) = \frac{1 + 145}{12}

sin (x) = - \frac{- 1 + 145}{12}, 0 \leq x < 2 π : x = π + arcsin (\frac{145 - 1}{12}), x = - arcsin (\frac{145 - 1}{12}) + 2 π

sin (x) = - \frac{- 1 + 145}{12}, 0 \leq x < 2 π

Apply trig inverse properties

sin (x) = - \frac{- 1 + 145}{12}

General solutions for

sin (x) = - \frac{- 1 + 145}{12}

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

x = arcsin (- \frac{- 1 + 145}{12}) + 2 πn, x = π + arcsin (\frac{- 1 + 145}{12}) + 2 πn

x = arcsin (- \frac{- 1 + 145}{12}) + 2 πn, x = π + arcsin (\frac{- 1 + 145}{12}) + 2 πn

Solutions for the range

0 \leq x < 2 π

x = π + arcsin (\frac{145 - 1}{12}), x = - arcsin (\frac{145 - 1}{12}) + 2 π

sin (x) = \frac{1 + 145}{12}, 0 \leq x < 2 π :

No Solution

sin (x) = \frac{1 + 145}{12}, 0 \leq x < 2 π

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = π + arcsin (\frac{145 - 1}{12}), x = - arcsin (\frac{145 - 1}{12}) + 2 π

6 cos^{2} (x) - sin (x) = 0, 0 \leq x < 2 π : x = arcsin (\frac{145 - 1}{12}), x = π - arcsin (\frac{145 - 1}{12})

6 cos^{2} (x) - sin (x) = 0, 0 \leq x < 2 π

Rewrite using trig identities

- sin (x) + 6 cos^{2} (x)

Use the Pythagorean identity:

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

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

= - sin (x) + 6 (1 - sin^{2} (x))

- sin (x) + (1 - sin^{2} (x)) \cdot 6 = 0

Solve by substitution

- sin (x) + (1 - sin^{2} (x)) \cdot 6 = 0

Let:

sin (x) = u

- u + (1 - u^{2}) \cdot 6 = 0

- u + (1 - u^{2}) \cdot 6 = 0 : u = - \frac{1 + 145}{12}, u = \frac{145 - 1}{12}

- u + (1 - u^{2}) \cdot 6 = 0

Expand

- u + (1 - u^{2}) \cdot 6 : - u + 6 - 6 u^{2}

- u + (1 - u^{2}) \cdot 6

= - u + 6 (1 - u^{2})

Expand

6 (1 - u^{2}) : 6 - 6 u^{2}

6 (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = 6, b = 1, c = u^{2}

= 6 \cdot 1 - 6 u^{2}

Multiply the numbers:

6 \cdot 1 = 6

= 6 - 6 u^{2}

= - u + 6 - 6 u^{2}

- u + 6 - 6 u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

- 6 u^{2} - u + 6 = 0

Solve with the quadratic formula

- 6 u^{2} - u + 6 = 0

Quadratic Equation Formula:

For

a = - 6, b = - 1, c = 6

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 6 ) \cdot 6}{2 ( - 6 )}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 6 ) \cdot 6}{2 ( - 6 )}

(- 1)^{2} - 4 (- 6) \cdot 6 = 145

(- 1)^{2} - 4 (- 6) \cdot 6

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 6 \cdot 6

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 6 \cdot 6 = 144

4 \cdot 6 \cdot 6

Multiply the numbers:

4 \cdot 6 \cdot 6 = 144

= 144

= 1 + 144

Add the numbers:

1 + 144 = 145

= 145

u_{1, 2} = \frac{- ( - 1 ) \pm 145}{2 ( - 6 )}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 145}{2 ( - 6 )}, u_{2} = \frac{- ( - 1 ) - 145}{2 ( - 6 )}

u = \frac{- ( - 1 ) + 145}{2 ( - 6 )} : - \frac{1 + 145}{12}

\frac{- ( - 1 ) + 145}{2 ( - 6 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{1 + 145}{- 2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{1 + 145}{- 12}

Apply the fraction rule:

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

= - \frac{1 + 145}{12}

u = \frac{- ( - 1 ) - 145}{2 ( - 6 )} : \frac{145 - 1}{12}

\frac{- ( - 1 ) - 145}{2 ( - 6 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{1 - 145}{- 2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{1 - 145}{- 12}

Apply the fraction rule:

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

1 - 145 = - (145 - 1)

= \frac{145 - 1}{12}

The solutions to the quadratic equation are:

u = - \frac{1 + 145}{12}, u = \frac{145 - 1}{12}

Substitute back

u = sin (x)

sin (x) = - \frac{1 + 145}{12}, sin (x) = \frac{145 - 1}{12}

sin (x) = - \frac{1 + 145}{12}, sin (x) = \frac{145 - 1}{12}

sin (x) = - \frac{1 + 145}{12}, 0 \leq x < 2 π :

No Solution

sin (x) = - \frac{1 + 145}{12}, 0 \leq x < 2 π

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = \frac{145 - 1}{12}, 0 \leq x < 2 π : x = arcsin (\frac{145 - 1}{12}), x = π - arcsin (\frac{145 - 1}{12})

sin (x) = \frac{145 - 1}{12}, 0 \leq x < 2 π

Apply trig inverse properties

sin (x) = \frac{145 - 1}{12}

General solutions for

sin (x) = \frac{145 - 1}{12}

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

x = arcsin (\frac{145 - 1}{12}) + 2 πn, x = π - arcsin (\frac{145 - 1}{12}) + 2 πn

x = arcsin (\frac{145 - 1}{12}) + 2 πn, x = π - arcsin (\frac{145 - 1}{12}) + 2 πn

Solutions for the range

0 \leq x < 2 π

x = arcsin (\frac{145 - 1}{12}), x = π - arcsin (\frac{145 - 1}{12})

Combine all the solutions

x = arcsin (\frac{145 - 1}{12}), x = π - arcsin (\frac{145 - 1}{12})

Combine all the solutions

x = π + arcsin (\frac{145 - 1}{12}), x = - arcsin (\frac{145 - 1}{12}) + 2 π, x = arcsin (\frac{145 - 1}{12}), x = π - arcsin (\frac{145 - 1}{12})

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

6 cos (x) - tan (x) = 0

Remove the ones that don't agree with the equation.

Check the solution

π + arcsin (\frac{145 - 1}{12}) :

False

π + arcsin (\frac{145 - 1}{12})

Plug in

n = 1

π + arcsin (\frac{145 - 1}{12})

For

6 cos (x) - tan (x) = 0

plug in

x = π + arcsin (\frac{145 - 1}{12})

6 cos (π + arcsin (\frac{145 - 1}{12})) - tan (π + arcsin (\frac{145 - 1}{12})) = 0

Refine

- 4.69927 \dots = 0

\Rightarrow False

Check the solution

- arcsin (\frac{145 - 1}{12}) + 2 π :

False

- arcsin (\frac{145 - 1}{12}) + 2 π

Plug in

n = 1

- arcsin (\frac{145 - 1}{12}) + 2 π

For

6 cos (x) - tan (x) = 0

plug in

x = - arcsin (\frac{145 - 1}{12}) + 2 π

6 cos (- arcsin (\frac{145 - 1}{12}) + 2 π) - tan (- arcsin (\frac{145 - 1}{12}) + 2 π) = 0

Refine

4.69927 \dots = 0

\Rightarrow False

Check the solution

arcsin (\frac{145 - 1}{12}) :

True

arcsin (\frac{145 - 1}{12})

Plug in

n = 1

arcsin (\frac{145 - 1}{12})

For

6 cos (x) - tan (x) = 0

plug in

x = arcsin (\frac{145 - 1}{12})

6 cos (arcsin (\frac{145 - 1}{12})) - tan (arcsin (\frac{145 - 1}{12})) = 0

Refine

0 = 0

\Rightarrow True

Check the solution

π - arcsin (\frac{145 - 1}{12}) :

True

π - arcsin (\frac{145 - 1}{12})

Plug in

n = 1

π - arcsin (\frac{145 - 1}{12})

For

6 cos (x) - tan (x) = 0

plug in

x = π - arcsin (\frac{145 - 1}{12})

6 cos (π - arcsin (\frac{145 - 1}{12})) - tan (π - arcsin (\frac{145 - 1}{12})) = 0

Refine

0 = 0

\Rightarrow True

x = arcsin (\frac{145 - 1}{12}), x = π - arcsin (\frac{145 - 1}{12})

Show solutions in decimal form

x = 1.16841 \dots, x = π - 1.16841 \dots

Combine all the solutions

1.16841 \dots or \frac{π}{2} or π - 1.16841 \dots or \frac{3 π}{2}

The intervals between the zeros

0 < x < 1.16841 \dots, 1.16841 \dots < x < \frac{π}{2}, \frac{π}{2} < x < π - 1.16841 \dots, π - 1.16841 \dots < x < \frac{3 π}{2}, \frac{3 π}{2} < x < 2 π

Summarize in a table:

cos (x) 6 cos (x) - tan (x) cos (x) (6 cos (x) - tan (x)) x = 0 + + + 0 < x < 1.16841 \dots + + + x = 1.16841 \dots + 00 1.16841 \dots < x < \frac{π}{2} + - - x = \frac{π}{2} 0 Undefined Undefined \frac{π}{2} < x < π - 1.16841 \dots - + - x = π - 1.16841 \dots - 00 π - 1.16841 \dots < x < \frac{3 π}{2} - - + x = \frac{3 π}{2} 0 Undefined Undefined \frac{3 π}{2} < x < 2 π + + + x = 2 π + + +

Identify the intervals that satisfy the required condition:

> 0

x = 0 or 0 < x < 1.16841 \dots or π - 1.16841 \dots < x < \frac{3 π}{2} or \frac{3 π}{2} < x < 2 π or x = 2 π

Merge Overlapping Intervals

0 \leq x < 1.16841 \dots or π - 1.16841 \dots < x < \frac{3 π}{2} or \frac{3 π}{2} < x < 2 π or x = 2 π

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

x = 0

0 < x < 1.16841 \dots

0 \leq x < 1.16841 \dots

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

0 \leq x < 1.16841 \dots

π - 1.16841 \dots < x < \frac{3 π}{2}

0 \leq x < 1.16841 \dots or π - 1.16841 \dots < x < \frac{3 π}{2}

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

0 \leq x < 1.16841 \dots or π - 1.16841 \dots < x < \frac{3 π}{2}

\frac{3 π}{2} < x < 2 π

0 \leq x < 1.16841 \dots or π - 1.16841 \dots < x < \frac{3 π}{2} or \frac{3 π}{2} < x < 2 π

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

0 \leq x < 1.16841 \dots or π - 1.16841 \dots < x < \frac{3 π}{2} or \frac{3 π}{2} < x < 2 π

x = 2 π

0 \leq x < 1.16841 \dots or π - 1.16841 \dots < x < \frac{3 π}{2} or \frac{3 π}{2} < x \leq 2 π

0 \leq x < 1.16841 \dots or π - 1.16841 \dots < x < \frac{3 π}{2} or \frac{3 π}{2} < x \leq 2 π

Apply the periodicity of

6 cos^{2} (x) - tan (x) cos (x)

2 πn \leq x < 1.16841 \dots + 2 πn or π - 1.16841 \dots + 2 πn < x < \frac{3 π}{2} + 2 πn or \frac{3 π}{2} + 2 πn < x \leq 2 π + 2 πn

Popular Examples

sin(x)*cos(x)>0

sin (x) \cdot cos (x) > 0

cos^2(x)+0.5>0

cos^{2} (x) + 0.5 > 0

cos(pix)> 1/2

cos (π x) > \frac{1}{2}

cos(θ)<0,sin(θ)<0

cos (θ) < 0, sin (θ) < 0

sin(5x)>= 5

sin (5 x) \geq 5