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Popular Trigonometry >

sin(x)<cos(x)<tan(x)

Solution

sin (x) < cos (x) < tan (x)

Solution

0.66623 \dots + 2 πn < x < \frac{π}{4} + 2 πn or \frac{5 π}{4} + 2 πn < x < \frac{3 π}{2} + 2 πn

Interval Notation

(0.66623 \dots + 2 πn, \frac{π}{4} + 2 πn) \cup (\frac{5 π}{4} + 2 πn, \frac{3 π}{2} + 2 πn)

Decimal

0.66623 \dots + 2 πn < x < 0.78539 \dots + 2 πn or 3.92699 \dots + 2 πn < x < 4.71238 \dots + 2 πn

Solution steps

sin (x) < cos (x) < tan (x)

a < u < b

then

a < u and u < b

sin (x) < cos (x) and cos (x) < tan (x)

sin (x) < cos (x) : - \frac{3 π}{4} + 2 πn < x < \frac{π}{4} + 2 πn

sin (x) < cos (x)

Move

cos (x)

to the left side

sin (x) < cos (x)

Subtract

cos (x)

from both sides

sin (x) - cos (x) < cos (x) - cos (x)

sin (x) - cos (x) < 0

sin (x) - cos (x) < 0

Use the following identity:

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

- 2 cos (\frac{π}{4} + x) < 0

Multiply both sides by

- 1

- 2 cos (\frac{π}{4} + x) < 0

Multiply both sides by -1 (reverse the inequality)

(- 2 cos (\frac{π}{4} + x)) (- 1) > 0 \cdot (- 1)

Simplify

2 cos (\frac{π}{4} + x) > 0

2 cos (\frac{π}{4} + x) > 0

Divide both sides by

2

2 cos (\frac{π}{4} + x) > 0

Divide both sides by

2

\frac{2 cos ( \frac{π}{4} + x )}{2} > \frac{0}{2}

Simplify

cos (\frac{π}{4} + x) > 0

cos (\frac{π}{4} + 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 < (\frac{π}{4} + x) < arccos (0) + 2 πn

a < u < b

then

a < u and u < b

- arccos (0) + 2 πn < \frac{π}{4} + x and \frac{π}{4} + x < arccos (0) + 2 πn

- arccos (0) + 2 πn < \frac{π}{4} + x : x > 2 πn - \frac{3 π}{4}

- arccos (0) + 2 πn < \frac{π}{4} + x

Switch sides

\frac{π}{4} + x > - arccos (0) + 2 πn

Simplify

- arccos (0) + 2 πn : - \frac{π}{2} + 2 πn

- arccos (0) + 2 πn

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} + 2 πn

\frac{π}{4} + x > - \frac{π}{2} + 2 πn

Move

\frac{π}{4}

to the right side

\frac{π}{4} + x > - \frac{π}{2} + 2 πn

Subtract

\frac{π}{4}

from both sides

\frac{π}{4} + x - \frac{π}{4} > - \frac{π}{2} + 2 πn - \frac{π}{4}

Simplify

\frac{π}{4} + x - \frac{π}{4} > - \frac{π}{2} + 2 πn - \frac{π}{4}

Simplify

\frac{π}{4} + x - \frac{π}{4} : x

\frac{π}{4} + x - \frac{π}{4}

Add similar elements:

\frac{π}{4} - \frac{π}{4} > 0

= x

Simplify

- \frac{π}{2} + 2 πn - \frac{π}{4} : 2 πn - \frac{3 π}{4}

- \frac{π}{2} + 2 πn - \frac{π}{4}

Group like terms

= 2 πn - \frac{π}{2} - \frac{π}{4}

Least Common Multiplier of

2, 4 : 4

2, 4

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

Multiply each factor the greatest number of times it occurs in either

2

4

= 2 \cdot 2

Multiply the numbers:

2 \cdot 2 = 4

= 4

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

4

For

\frac{π}{2} :

multiply the denominator and numerator by

2

\frac{π}{2} = \frac{π 2}{2 \cdot 2} = \frac{π 2}{4}

= - \frac{π 2}{4} - \frac{π}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{- π 2 - π}{4}

Add similar elements:

- 2 π - π = - 3 π

= \frac{- 3 π}{4}

Apply the fraction rule:

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

= 2 πn - \frac{3 π}{4}

x > 2 πn - \frac{3 π}{4}

x > 2 πn - \frac{3 π}{4}

x > 2 πn - \frac{3 π}{4}

\frac{π}{4} + x < arccos (0) + 2 πn : x < 2 πn + \frac{π}{4}

\frac{π}{4} + x < arccos (0) + 2 πn

Simplify

arccos (0) + 2 πn : \frac{π}{2} + 2 πn

arccos (0) + 2 πn

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} + 2 πn

\frac{π}{4} + x < \frac{π}{2} + 2 πn

Move

\frac{π}{4}

to the right side

\frac{π}{4} + x < \frac{π}{2} + 2 πn

Subtract

\frac{π}{4}

from both sides

\frac{π}{4} + x - \frac{π}{4} < \frac{π}{2} + 2 πn - \frac{π}{4}

Simplify

\frac{π}{4} + x - \frac{π}{4} < \frac{π}{2} + 2 πn - \frac{π}{4}

Simplify

\frac{π}{4} + x - \frac{π}{4} : x

\frac{π}{4} + x - \frac{π}{4}

Add similar elements:

\frac{π}{4} - \frac{π}{4} < 0

= x

Simplify

\frac{π}{2} + 2 πn - \frac{π}{4} : 2 πn + \frac{π}{4}

\frac{π}{2} + 2 πn - \frac{π}{4}

Group like terms

= 2 πn + \frac{π}{2} - \frac{π}{4}

Least Common Multiplier of

2, 4 : 4

2, 4

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

Multiply each factor the greatest number of times it occurs in either

2

4

= 2 \cdot 2

Multiply the numbers:

2 \cdot 2 = 4

= 4

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

4

For

\frac{π}{2} :

multiply the denominator and numerator by

2

\frac{π}{2} = \frac{π 2}{2 \cdot 2} = \frac{π 2}{4}

= \frac{π 2}{4} - \frac{π}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{π 2 - π}{4}

Add similar elements:

2 π - π = π

= 2 πn + \frac{π}{4}

x < 2 πn + \frac{π}{4}

x < 2 πn + \frac{π}{4}

x < 2 πn + \frac{π}{4}

Combine the intervals

x > 2 πn - \frac{3 π}{4} and x < 2 πn + \frac{π}{4}

Merge Overlapping Intervals

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

cos (x) < tan (x) : 0.66623 \dots + 2 πn < x < \frac{π}{2} + 2 πn or π - 0.66623 \dots + 2 πn < x < \frac{3 π}{2} + 2 πn

cos (x) < tan (x)

Move

tan (x)

to the left side

cos (x) < tan (x)

Subtract

tan (x)

from both sides

cos (x) - tan (x) < tan (x) - tan (x)

cos (x) - tan (x) < 0

cos (x) - tan (x) < 0

Periodicity of

cos (x) - tan (x) : 2 π

The compound periodicity of the sum of periodic functions is the least common multiplier of the periods

cos (x), tan (x)

Periodicity of

cos (x) : 2 π

Periodicity of

cos (x)

2 π

= 2 π

Periodicity of

tan (x) : π

Periodicity of

tan (x)

π

= π

Combine periods:

2 π, π

= 2 π

Express with sin, cos

cos (x) - tan (x) < 0

Use the basic trigonometric identity:

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

cos (x) - \frac{sin ( x )}{cos ( x )} < 0

cos (x) - \frac{sin ( x )}{cos ( x )} < 0

Simplify

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

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

Convert element to fraction:

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

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

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

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

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

cos (x) cos (x)

Apply exponent rule:

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

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

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

Find the zeroes and undifined points of

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

for

0 \leq x < 2 π

To find the zeroes, set the inequality to zero

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

\frac{cos ^{2} ( x ) - sin ( x )}{cos ( x )} = 0, 0 \leq x < 2 π : x = 0.66623 \dots, x = π - 0.66623 \dots

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Solve by substitution

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

Let:

sin (x) = u

1 - u - u^{2} = 0

1 - u - u^{2} = 0 : u = - \frac{1 + 5}{2}, u = \frac{5 - 1}{2}

1 - u - u^{2} = 0

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

(- 1)^{2} - 4 (- 1) \cdot 1 = 5

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

Apply rule

- (- a) = a

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

(- 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 1 \cdot 1 = 4

4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 4

= 1 + 4

Add the numbers:

1 + 4 = 5

= 5

u_{1, 2} = \frac{- ( - 1 ) \pm 5}{2 ( - 1 )}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 5}{2 ( - 1 )}, u_{2} = \frac{- ( - 1 ) - 5}{2 ( - 1 )}

u = \frac{- ( - 1 ) + 5}{2 ( - 1 )} : - \frac{1 + 5}{2}

\frac{- ( - 1 ) + 5}{2 ( - 1 )}

Remove parentheses:

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

= \frac{1 + 5}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{1 + 5}{- 2}

Apply the fraction rule:

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

= - \frac{1 + 5}{2}

u = \frac{- ( - 1 ) - 5}{2 ( - 1 )} : \frac{5 - 1}{2}

\frac{- ( - 1 ) - 5}{2 ( - 1 )}

Remove parentheses:

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

= \frac{1 - 5}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{1 - 5}{- 2}

Apply the fraction rule:

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

1 - 5 = - (5 - 1)

= \frac{5 - 1}{2}

The solutions to the quadratic equation are:

u = - \frac{1 + 5}{2}, u = \frac{5 - 1}{2}

Substitute back

u = sin (x)

sin (x) = - \frac{1 + 5}{2}, sin (x) = \frac{5 - 1}{2}

sin (x) = - \frac{1 + 5}{2}, sin (x) = \frac{5 - 1}{2}

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

No Solution

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

- 1 \leq sin (x) \leq 1

No Solution

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

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

Apply trig inverse properties

sin (x) = \frac{5 - 1}{2}

General solutions for

sin (x) = \frac{5 - 1}{2}

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

x = arcsin (\frac{5 - 1}{2}) + 2 πn, x = π - arcsin (\frac{5 - 1}{2}) + 2 πn

x = arcsin (\frac{5 - 1}{2}) + 2 πn, x = π - arcsin (\frac{5 - 1}{2}) + 2 πn

Solutions for the range

0 \leq x < 2 π

x = arcsin (\frac{5 - 1}{2}), x = π - arcsin (\frac{5 - 1}{2})

Combine all the solutions

x = arcsin (\frac{5 - 1}{2}), x = π - arcsin (\frac{5 - 1}{2})

Show solutions in decimal form

x = 0.66623 \dots, x = π - 0.66623 \dots

Find the undefined points:

x = \frac{π}{2}, x = \frac{3 π}{2}

Find the zeros of the denominator

cos (x) = 0

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}

0.66623 \dots, \frac{π}{2}, π - 0.66623 \dots, \frac{3 π}{2}

Identify the intervals

0 < x < 0.66623 \dots, 0.66623 \dots < x < \frac{π}{2}, \frac{π}{2} < x < π - 0.66623 \dots, π - 0.66623 \dots < x < \frac{3 π}{2}, \frac{3 π}{2} < x < 2 π

Summarize in a table:

cos^{2} (x) - sin (x) cos (x) \frac{c o s ^{2} ( x ) - s i n ( x )}{c o s ( x )} x = 0 + + + 0 < x < 0.66623 \dots + + + x = 0.66623 \dots 0 + 0 0.66623 \dots < x < \frac{π}{2} - + - x = \frac{π}{2} - 0 Undefined \frac{π}{2} < x < π - 0.66623 \dots - - + x = π - 0.66623 \dots 0 - 0 π - 0.66623 \dots < x < \frac{3 π}{2} + - - x = \frac{3 π}{2} + 0 Undefined \frac{3 π}{2} < x < 2 π + + + x = 2 π + + +

Identify the intervals that satisfy the required condition:

< 0

0.66623 \dots < x < \frac{π}{2} or π - 0.66623 \dots < x < \frac{3 π}{2}

Apply the periodicity of

cos (x) - tan (x)

0.66623 \dots + 2 πn < x < \frac{π}{2} + 2 πn or π - 0.66623 \dots + 2 πn < x < \frac{3 π}{2} + 2 πn

Combine the intervals

- \frac{3 π}{4} + 2 πn < x < \frac{π}{4} + 2 πn and (0.66623 \dots + 2 πn < x < \frac{π}{2} + 2 πn or π - 0.66623 \dots + 2 πn < x < \frac{3 π}{2} + 2 πn)

Merge Overlapping Intervals

0.66623 \dots + 2 πn < x < \frac{π}{4} + 2 πn or \frac{5 π}{4} + 2 πn < x < \frac{3 π}{2} + 2 πn

Popular Examples

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sin (x) 0 \leq x \leq 2 π

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sin(2x)>= 0\land cos(x)>0

sin (2 x) \geq 0 and cos (x) > 0

cos(x)= 5/13 \land sin(x)<0,tan(2x)

cos (x) = \frac{5}{13} and sin (x) < 0, tan (2 x)

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