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sec^2(x)<= tan^2(x)+sec(x)

Solution

sec^{2} (x) \leq tan^{2} (x) + sec (x)

Solution

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

Interval Notation

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

Decimal Notation

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

Solution steps

sec^{2} (x) \leq tan^{2} (x) + sec (x)

Subtract

tan^{2} (x) + sec (x)

from both sides

sec^{2} (x) - (tan^{2} (x) + sec (x)) \leq tan^{2} (x) + sec (x) - (tan^{2} (x) + sec (x))

sec^{2} (x) - (tan^{2} (x) + sec (x)) \leq 0

Periodicity of

sec^{2} (x) - (tan^{2} (x) + sec (x)) : 2 π

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

sec^{2} (x), (tan^{2} (x) + sec (x))

Periodicity of

sec^{2} (x) : π

Periodicity of

sec^{n} (x) = \frac{Periodicity of sec ( x )}{2},

if n is even

Periodicity of

sec (x) : 2 π

Periodicity of

sec (x)

2 π

= 2 π

\frac{2 π}{2}

Simplify

π

Periodicity of

(tan^{2} (x) + sec (x)) : 2 π

(tan^{2} (x) + sec (x))

is composed of the following functions and periods:

sec (x)

with periodicity of

2 π

The compound periodicity is:

2 π

Combine periods:

π, 2 π

= 2 π

Express with sin, cos

sec^{2} (x) - (tan^{2} (x) + sec (x)) \leq 0

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

(\frac{1}{cos ( x )})^{2} - (tan^{2} (x) + \frac{1}{cos ( x )}) \leq 0

Use the basic trigonometric identity:

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

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

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

Simplify

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

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

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

(\frac{1}{cos ( x )})^{2}

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

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

Apply exponent rule:

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

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

Join

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

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

Least Common Multiplier of

cos^{2} (x), cos (x) : cos^{2} (x)

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

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

cos^{2} (x)

cos (x)

= cos^{2} (x)

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

cos^{2} (x)

For

\frac{1}{cos ( x )} :

multiply the denominator and numerator by

cos (x)

\frac{1}{cos ( x )} = \frac{1 \cdot cos ( x )}{cos ( x ) cos ( x )} = \frac{cos ( x )}{cos ^{2} ( x )}

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

Since the denominators are equal, combine the fractions:

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

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

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

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

Apply rule

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

+ (- a) = - a

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

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

\frac{1 - sin ^{2} ( x ) - cos ( x )}{cos ^{2} ( x )} \leq 0

Find the zeroes and undifined points of

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

for

0 \leq x < 2 π

To find the zeroes, set the inequality to zero

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

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

- u + u^{2} = 0

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

- u + u^{2} = 0

Write in the standard form

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

u^{2} - u = 0

Solve with the quadratic formula

u^{2} - u = 0

Quadratic Equation Formula:

For

a = 1, b = - 1, c = 0

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot 0 = 1

(- 1)^{2} - 4 \cdot 1 \cdot 0

(- 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 0 = 0

4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 0

= 1 - 0

Subtract the numbers:

1 - 0 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

1 + 1 = 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 1 ) - 1}{2 \cdot 1} : 0

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

Apply rule

- (- a) = a

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

Subtract the numbers:

1 - 1 = 0

= \frac{0}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

The solutions to the quadratic equation are:

u = 1, u = 0

Substitute back

u = cos (x)

cos (x) = 1, cos (x) = 0

cos (x) = 1, cos (x) = 0

cos (x) = 1, 0 \leq x < 2 π : x = 0

cos (x) = 1, 0 \leq x < 2 π

General solutions for

cos (x) = 1

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 = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

Solutions for the range

0 \leq x < 2 π

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}

Combine all the solutions

x = 0, x = \frac{π}{2}, x = \frac{3 π}{2}

Since the equation is undefined for:

\frac{π}{2}, \frac{3 π}{2}

x = 0

Find the undefined points:

x = \frac{π}{2}, x = \frac{3 π}{2}

Find the zeros of the denominator

cos^{2} (x) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

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, \frac{π}{2}, \frac{3 π}{2}

Identify the intervals

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

Summarize in a table:

1 - sin^{2} (x) - cos (x) cos^{2} (x) \frac{1 - s i n ^{2} ( x ) - c o s ( x )}{c o s ^{2} ( x )} x = 0 0 + 0 0 < x < \frac{π}{2} - + - x = \frac{π}{2} 00 Undefined \frac{π}{2} < x < \frac{3 π}{2} + + + x = \frac{3 π}{2} 00 Undefined \frac{3 π}{2} < x < 2 π - + - x = 2 π 0 + 0

Identify the intervals that satisfy the required condition:

\leq 0

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

Merge Overlapping Intervals

0 \leq x < \frac{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 < \frac{π}{2}

0 \leq x < \frac{π}{2}

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

0 \leq x < \frac{π}{2}

\frac{3 π}{2} < x < 2 π

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

x = 2 π

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

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

Apply the periodicity of

sec^{2} (x) - (tan^{2} (x) + sec (x))

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

Popular Examples

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sin (x) - 3 cos (x) \leq 0

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

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

cos(t)>= 0

cos (t) \geq 0

-1+tan(x)<= 1

- 1 + tan (x) \leq 1

cos(x)(2sin(x)-1)<= 0,-pi<x<= pi

cos (x) (2 sin (x) - 1) \leq 0, - π < x \leq π