English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sec^2(θ)tan(θ)>= 2tan(θ)

Solution

sec^{2} (θ) tan (θ) \geq 2 tan (θ)

Solution

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

Interval Notation

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

Decimal Notation

0.78539 \dots + πn \leq θ < 1.57079 \dots + πn or 2.35619 \dots + πn \leq θ \leq 3.14159 \dots + πn

Solution steps

sec^{2} (θ) tan (θ) \geq 2 tan (θ)

Move

2 tan (θ)

to the left side

sec^{2} (θ) tan (θ) \geq 2 tan (θ)

Subtract

2 tan (θ)

from both sides

sec^{2} (θ) tan (θ) - 2 tan (θ) \geq 2 tan (θ) - 2 tan (θ)

sec^{2} (θ) tan (θ) - 2 tan (θ) \geq 0

sec^{2} (θ) tan (θ) - 2 tan (θ) \geq 0

Periodicity of

sec^{2} (θ) tan (θ) - 2 tan (θ) : π

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

sec^{2} (θ) tan (θ), 2 tan (θ)

Periodicity of

sec^{2} (θ) tan (θ) : π

sec^{2} (θ) tan (θ)

is composed of the following functions and periods:

sec (θ)

with periodicity of

2 π

The compound periodicity is:

π

Periodicity of

2 tan (θ) : π

Periodicity of

a \cdot tan (b x + c) + d = \frac{periodicity of tan ( x )}{∣ b ∣}

Periodicity of

tan (x)

π

= \frac{π}{∣ 1 ∣}

Simplify

= π

Combine periods:

π, π

= π

Express with sin, cos

sec^{2} (θ) tan (θ) - 2 tan (θ) \geq 0

Use the basic trigonometric identity:

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

(\frac{1}{cos ( θ )})^{2} tan (θ) - 2 tan (θ) \geq 0

Use the basic trigonometric identity:

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

(\frac{1}{cos ( θ )})^{2} \frac{sin ( θ )}{cos ( θ )} - 2 \cdot \frac{sin ( θ )}{cos ( θ )} \geq 0

(\frac{1}{cos ( θ )})^{2} \frac{sin ( θ )}{cos ( θ )} - 2 \cdot \frac{sin ( θ )}{cos ( θ )} \geq 0

Simplify

(\frac{1}{cos ( θ )})^{2} \frac{sin ( θ )}{cos ( θ )} - 2 \cdot \frac{sin ( θ )}{cos ( θ )} : \frac{sin ( θ ) - 2 cos ^{2} ( θ ) sin ( θ )}{cos ^{3} ( θ )}

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

(\frac{1}{cos ( θ )})^{2} \frac{sin ( θ )}{cos ( θ )} = \frac{sin ( θ )}{cos ^{3} ( θ )}

(\frac{1}{cos ( θ )})^{2} \frac{sin ( θ )}{cos ( θ )}

Multiply fractions:

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

= \frac{sin ( θ ) ( \frac{1}{c o s ( θ )} ) ^{2}}{cos ( θ )}

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

(\frac{1}{cos ( θ )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{cos ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( θ )}

= \frac{\frac{1}{c o s ^{2} ( θ )} sin ( θ )}{cos ( θ )}

Multiply

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

sin (θ) \frac{1}{cos ^{2} ( θ )}

Multiply fractions:

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

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

Multiply:

1 \cdot sin (θ) = sin (θ)

= \frac{sin ( θ )}{cos ^{2} ( θ )}

= \frac{\frac{s i n ( θ )}{c o s ^{2} ( θ )}}{cos ( θ )}

Apply the fraction rule:

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

= \frac{sin ( θ )}{cos ^{2} ( θ ) cos ( θ )}

cos^{2} (θ) cos (θ) = cos^{3} (θ)

cos^{2} (θ) cos (θ)

Apply exponent rule:

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

cos^{2} (θ) cos (θ) = cos^{2 + 1} (θ)

= cos^{2 + 1} (θ)

Add the numbers:

2 + 1 = 3

= cos^{3} (θ)

= \frac{sin ( θ )}{cos ^{3} ( θ )}

2 \cdot \frac{sin ( θ )}{cos ( θ )} = \frac{2 sin ( θ )}{cos ( θ )}

2 \cdot \frac{sin ( θ )}{cos ( θ )}

Multiply fractions:

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

= \frac{sin ( θ ) \cdot 2}{cos ( θ )}

= \frac{sin ( θ )}{cos ^{3} ( θ )} - \frac{2 sin ( θ )}{cos ( θ )}

Least Common Multiplier of

cos^{3} (θ), cos (θ) : cos^{3} (θ)

cos^{3} (θ), cos (θ)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

cos^{3} (θ)

cos (θ)

= cos^{3} (θ)

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^{3} (θ)

For

\frac{sin ( θ ) \cdot 2}{cos ( θ )} :

multiply the denominator and numerator by

cos^{2} (θ)

\frac{sin ( θ ) \cdot 2}{cos ( θ )} = \frac{sin ( θ ) \cdot 2 cos ^{2} ( θ )}{cos ( θ ) cos ^{2} ( θ )} = \frac{sin ( θ ) \cdot 2 cos ^{2} ( θ )}{cos ^{3} ( θ )}

= \frac{sin ( θ )}{cos ^{3} ( θ )} - \frac{sin ( θ ) \cdot 2 cos ^{2} ( θ )}{cos ^{3} ( θ )}

Since the denominators are equal, combine the fractions:

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

= \frac{sin ( θ ) - sin ( θ ) \cdot 2 cos ^{2} ( θ )}{cos ^{3} ( θ )}

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

Find the zeroes and undifined points of

\frac{sin ( θ ) - 2 cos ^{2} ( θ ) sin ( θ )}{cos ^{3} ( θ )}

for

0 \leq θ < π

To find the zeroes, set the inequality to zero

\frac{sin ( θ ) - 2 cos ^{2} ( θ ) sin ( θ )}{cos ^{3} ( θ )} = 0

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

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

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

sin (θ) - 2 cos^{2} (θ) sin (θ) = 0

Factor

sin (θ) - 2 cos^{2} (θ) sin (θ) : - sin (θ) (2 cos (θ) + 1) (2 cos (θ) - 1)

sin (θ) - 2 cos^{2} (θ) sin (θ)

Factor out common term

- sin (θ)

= - sin (θ) (- 1 + 2 cos^{2} (θ))

Factor

2 cos^{2} (θ) - 1 : (2 cos (θ) + 1) (2 cos (θ) - 1)

2 cos^{2} (θ) - 1

Rewrite

2 cos^{2} (θ) - 1

(2 cos (θ))^{2} - 1^{2}

2 cos^{2} (θ) - 1

Apply radical rule:

a = (a)^{2}

2 = (2)^{2}

= (2)^{2} cos^{2} (θ) - 1

Rewrite

1

1^{2}

= (2)^{2} cos^{2} (θ) - 1^{2}

Apply exponent rule:

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

(2)^{2} cos^{2} (θ) = (2 cos (θ))^{2}

= (2 cos (θ))^{2} - 1^{2}

= (2 cos (θ))^{2} - 1^{2}

Apply Difference of Two Squares Formula:

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

(2 cos (θ))^{2} - 1^{2} = (2 cos (θ) + 1) (2 cos (θ) - 1)

= (2 cos (θ) + 1) (2 cos (θ) - 1)

= - sin (θ) (2 cos (θ) + 1) (2 cos (θ) - 1)

- sin (θ) (2 cos (θ) + 1) (2 cos (θ) - 1) = 0

Solving each part separately

sin (θ) = 0 or 2 cos (θ) + 1 = 0 or 2 cos (θ) - 1 = 0

sin (θ) = 0, 0 \leq θ < π : θ = 0

sin (θ) = 0, 0 \leq θ < π

General solutions for

sin (θ) = 0

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}

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

Solutions for the range

0 \leq θ < π

θ = 0

2 cos (θ) + 1 = 0, 0 \leq θ < π : θ = \frac{3 π}{4}

2 cos (θ) + 1 = 0, 0 \leq θ < π

Move

1

to the right side

2 cos (θ) + 1 = 0

Subtract

1

from both sides

2 cos (θ) + 1 - 1 = 0 - 1

Simplify

2 cos (θ) = - 1

2 cos (θ) = - 1

Divide both sides by

2

2 cos (θ) = - 1

Divide both sides by

2

\frac{2 cos ( θ )}{2} = \frac{- 1}{2}

Simplify

\frac{2 cos ( θ )}{2} = \frac{- 1}{2}

Simplify

\frac{2 cos ( θ )}{2} : cos (θ)

\frac{2 cos ( θ )}{2}

Cancel the common factor:

2

= cos (θ)

Simplify

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

\frac{- 1}{2}

Apply the fraction rule:

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

= - \frac{1}{2}

Rationalize

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

- \frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2}

cos (θ) = - \frac{2}{2}

cos (θ) = - \frac{2}{2}

cos (θ) = - \frac{2}{2}

General solutions for

cos (θ) = - \frac{2}{2}

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}

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

Solutions for the range

0 \leq θ < π

θ = \frac{3 π}{4}

2 cos (θ) - 1 = 0, 0 \leq θ < π : θ = \frac{π}{4}

2 cos (θ) - 1 = 0, 0 \leq θ < π

Move

1

to the right side

2 cos (θ) - 1 = 0

Add

1

to both sides

2 cos (θ) - 1 + 1 = 0 + 1

Simplify

2 cos (θ) = 1

2 cos (θ) = 1

Divide both sides by

2

2 cos (θ) = 1

Divide both sides by

2

\frac{2 cos ( θ )}{2} = \frac{1}{2}

Simplify

\frac{2 cos ( θ )}{2} = \frac{1}{2}

Simplify

\frac{2 cos ( θ )}{2} : cos (θ)

\frac{2 cos ( θ )}{2}

Cancel the common factor:

2

= cos (θ)

Simplify

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

\frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2}{2}

cos (θ) = \frac{2}{2}

cos (θ) = \frac{2}{2}

cos (θ) = \frac{2}{2}

General solutions for

cos (θ) = \frac{2}{2}

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}

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

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

Solutions for the range

0 \leq θ < π

θ = \frac{π}{4}

Combine all the solutions

θ = 0, θ = \frac{3 π}{4}, θ = \frac{π}{4}

Find the undefined points:

θ = \frac{π}{2}

Find the zeros of the denominator

cos^{3} (θ) = 0

Apply rule

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

cos (θ) = 0

General solutions for

cos (θ) = 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}

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Solutions for the range

0 \leq θ < π

θ = \frac{π}{2}

0, \frac{π}{4}, \frac{π}{2}, \frac{3 π}{4}

Identify the intervals

0 < θ < \frac{π}{4}, \frac{π}{4} < θ < \frac{π}{2}, \frac{π}{2} < θ < \frac{3 π}{4}, \frac{3 π}{4} < θ < π

Summarize in a table:

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

Identify the intervals that satisfy the required condition:

\geq 0

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

Merge Overlapping Intervals

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