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Popular Trigonometry >

tan(θ)-4/5 cos(θ)>0csc(θ)

Solution

tan (θ) - \frac{4}{5} cos (θ) > 0 csc (θ)

Solution

0.58745 \dots + 2 πn < θ < \frac{π}{2} + 2 πn or π - 0.58745 \dots + 2 πn < θ < \frac{3 π}{2} + 2 πn

Interval Notation

(0.58745 \dots + 2 πn, \frac{π}{2} + 2 πn) \cup (π - 0.58745 \dots + 2 πn, \frac{3 π}{2} + 2 πn)

Decimal

0.58745 \dots + 2 πn < θ < 1.57079 \dots + 2 πn or 2.55413 \dots + 2 πn < θ < 4.71238 \dots + 2 πn

Solution steps

tan (θ) - \frac{4}{5} cos (θ) > 0 \cdot csc (θ)

Move

0 csc (θ)

to the left side

tan (θ) - \frac{4}{5} cos (θ) > 0 \cdot csc (θ)

Subtract

0 csc (θ)

from both sides

tan (θ) - \frac{4}{5} cos (θ) - 0 \cdot csc (θ) > 0 \cdot csc (θ) - 0 \cdot csc (θ)

tan (θ) - \frac{4}{5} cos (θ) - 0 \cdot csc (θ) > 0 \cdot csc (θ) - 0 \cdot csc (θ)

Refine

Simplify

tan (θ) - \frac{4}{5} cos (θ) - 0 \cdot csc (θ) : tan (θ) - \frac{4}{5} cos (θ)

tan (θ) - \frac{4}{5} cos (θ) - 0 \cdot csc (θ)

Apply rule

0 \cdot a = 0

= tan (θ) - \frac{4}{5} cos (θ) - 0

tan (θ) - \frac{4}{5} cos (θ) - 0 = tan (θ) - \frac{4}{5} cos (θ)

= tan (θ) - \frac{4}{5} cos (θ)

0 \cdot csc (θ) - 0 \cdot csc (θ)

Add similar elements:

0 csc (θ) - 0 csc (θ) > 0

= 0

tan (θ) - \frac{4}{5} cos (θ) > 0

tan (θ) - \frac{4}{5} cos (θ) > 0

tan (θ) - \frac{4}{5} cos (θ) > 0

Periodicity of

tan (θ) - \frac{4}{5} cos (θ) : 2 π

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

tan (θ), \frac{4}{5} cos (θ)

Periodicity of

tan (θ) : π

Periodicity of

tan (x)

π

= π

Periodicity of

\frac{4}{5} cos (θ) : 2 π

Periodicity of

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

Periodicity of

cos (x)

2 π

= \frac{2 π}{∣ 1 ∣}

Simplify

= 2 π

Combine periods:

π, 2 π

= 2 π

Express with sin, cos

tan (θ) - \frac{4}{5} cos (θ) > 0

Use the basic trigonometric identity:

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

\frac{sin ( θ )}{cos ( θ )} - \frac{4}{5} cos (θ) > 0

\frac{sin ( θ )}{cos ( θ )} - \frac{4}{5} cos (θ) > 0

Simplify

\frac{sin ( θ )}{cos ( θ )} - \frac{4}{5} cos (θ) : \frac{5 sin ( θ ) - 4 cos ^{2} ( θ )}{5 cos ( θ )}

\frac{sin ( θ )}{cos ( θ )} - \frac{4}{5} cos (θ)

Multiply

\frac{4}{5} cos (θ) : \frac{4 cos ( θ )}{5}

\frac{4}{5} cos (θ)

Multiply fractions:

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

= \frac{4 cos ( θ )}{5}

= \frac{sin ( θ )}{cos ( θ )} - \frac{4 cos ( θ )}{5}

Least Common Multiplier of

cos (θ), 5 : 5 cos (θ)

cos (θ), 5

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

cos (θ)

5

= 5 cos (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

5 cos (θ)

For

\frac{sin ( θ )}{cos ( θ )} :

multiply the denominator and numerator by

5

\frac{sin ( θ )}{cos ( θ )} = \frac{sin ( θ ) \cdot 5}{cos ( θ ) \cdot 5}

For

\frac{4 cos ( θ )}{5} :

multiply the denominator and numerator by

cos (θ)

\frac{4 cos ( θ )}{5} = \frac{4 cos ( θ ) cos ( θ )}{5 cos ( θ )} = \frac{4 cos ^{2} ( θ )}{5 cos ( θ )}

= \frac{sin ( θ ) \cdot 5}{cos ( θ ) \cdot 5} - \frac{4 cos ^{2} ( θ )}{5 cos ( θ )}

Since the denominators are equal, combine the fractions:

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

= \frac{sin ( θ ) \cdot 5 - 4 cos ^{2} ( θ )}{5 cos ( θ )}

\frac{5 sin ( θ ) - 4 cos ^{2} ( θ )}{5 cos ( θ )} > 0

Find the zeroes and undifined points of

\frac{5 sin ( θ ) - 4 cos ^{2} ( θ )}{5 cos ( θ )}

for

0 \leq θ < 2 π

To find the zeroes, set the inequality to zero

\frac{5 sin ( θ ) - 4 cos ^{2} ( θ )}{5 cos ( θ )} = 0

\frac{5 sin ( θ ) - 4 cos ^{2} ( θ )}{5 cos ( θ )} = 0, 0 \leq θ < 2 π : θ = 0.58745 \dots, θ = π - 0.58745 \dots

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

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

5 sin (θ) - 4 cos^{2} (θ) = 0

Rewrite using trig identities

- 4 cos^{2} (θ) + 5 sin (θ)

Use the Pythagorean identity:

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

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

= - 4 (1 - sin^{2} (θ)) + 5 sin (θ)

- (1 - sin^{2} (θ)) \cdot 4 + 5 sin (θ) = 0

Solve by substitution

- (1 - sin^{2} (θ)) \cdot 4 + 5 sin (θ) = 0

Let:

sin (θ) = u

- (1 - u^{2}) \cdot 4 + 5 u = 0

- (1 - u^{2}) \cdot 4 + 5 u = 0 : u = \frac{- 5 + 89}{8}, u = \frac{- 5 - 89}{8}

- (1 - u^{2}) \cdot 4 + 5 u = 0

Expand

- (1 - u^{2}) \cdot 4 + 5 u : - 4 + 4 u^{2} + 5 u

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

= - 4 (1 - u^{2}) + 5 u

Expand

- 4 (1 - u^{2}) : - 4 + 4 u^{2}

- 4 (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = - 4, b = 1, c = u^{2}

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

Apply minus-plus rules

- (- a) = a

= - 4 \cdot 1 + 4 u^{2}

Multiply the numbers:

4 \cdot 1 = 4

= - 4 + 4 u^{2}

= - 4 + 4 u^{2} + 5 u

- 4 + 4 u^{2} + 5 u = 0

Write in the standard form

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

4 u^{2} + 5 u - 4 = 0

Solve with the quadratic formula

4 u^{2} + 5 u - 4 = 0

Quadratic Equation Formula:

For

a = 4, b = 5, c = - 4

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 4 ( - 4 )}{2 \cdot 4}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 4 ( - 4 )}{2 \cdot 4}

5^{2} - 4 \cdot 4 (- 4) = 89

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

Apply rule

- (- a) = a

= 5^{2} + 4 \cdot 4 \cdot 4

Multiply the numbers:

4 \cdot 4 \cdot 4 = 64

= 5^{2} + 64

5^{2} = 25

= 25 + 64

Add the numbers:

25 + 64 = 89

= 89

u_{1, 2} = \frac{- 5 \pm 89}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- 5 + 89}{2 \cdot 4}, u_{2} = \frac{- 5 - 89}{2 \cdot 4}

u = \frac{- 5 + 89}{2 \cdot 4} : \frac{- 5 + 89}{8}

\frac{- 5 + 89}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 5 + 89}{8}

u = \frac{- 5 - 89}{2 \cdot 4} : \frac{- 5 - 89}{8}

\frac{- 5 - 89}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 5 - 89}{8}

The solutions to the quadratic equation are:

u = \frac{- 5 + 89}{8}, u = \frac{- 5 - 89}{8}

Substitute back

u = sin (θ)

sin (θ) = \frac{- 5 + 89}{8}, sin (θ) = \frac{- 5 - 89}{8}

sin (θ) = \frac{- 5 + 89}{8}, sin (θ) = \frac{- 5 - 89}{8}

sin (θ) = \frac{- 5 + 89}{8}, 0 \leq θ < 2 π : θ = arcsin (\frac{89 - 5}{8}), θ = π - arcsin (\frac{89 - 5}{8})

sin (θ) = \frac{- 5 + 89}{8}, 0 \leq θ < 2 π

Apply trig inverse properties

sin (θ) = \frac{- 5 + 89}{8}

General solutions for

sin (θ) = \frac{- 5 + 89}{8}

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

θ = arcsin (\frac{- 5 + 89}{8}) + 2 πn, θ = π - arcsin (\frac{- 5 + 89}{8}) + 2 πn

θ = arcsin (\frac{- 5 + 89}{8}) + 2 πn, θ = π - arcsin (\frac{- 5 + 89}{8}) + 2 πn

Solutions for the range

0 \leq θ < 2 π

θ = arcsin (\frac{89 - 5}{8}), θ = π - arcsin (\frac{89 - 5}{8})

sin (θ) = \frac{- 5 - 89}{8}, 0 \leq θ < 2 π :

No Solution

sin (θ) = \frac{- 5 - 89}{8}, 0 \leq θ < 2 π

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

θ = arcsin (\frac{89 - 5}{8}), θ = π - arcsin (\frac{89 - 5}{8})

Show solutions in decimal form

θ = 0.58745 \dots, θ = π - 0.58745 \dots

Find the undefined points:

θ = \frac{π}{2}, θ = \frac{3 π}{2}

Find the zeros of the denominator

5 cos (θ) = 0

Divide both sides by

5

5 cos (θ) = 0

Divide both sides by

5

\frac{5 cos ( θ )}{5} = \frac{0}{5}

Simplify

cos (θ) = 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 < 2 π

θ = \frac{π}{2}, θ = \frac{3 π}{2}

0.58745 \dots, \frac{π}{2}, π - 0.58745 \dots, \frac{3 π}{2}

Identify the intervals

0 < θ < 0.58745 \dots, 0.58745 \dots < θ < \frac{π}{2}, \frac{π}{2} < θ < π - 0.58745 \dots, π - 0.58745 \dots < θ < \frac{3 π}{2}, \frac{3 π}{2} < θ < 2 π

Summarize in a table:

5 sin (θ) - 4 cos^{2} (θ) cos (θ) \frac{5 s i n ( θ ) - 4 c o s ^{2} ( θ )}{5 c o s ( θ )} θ = 0 - + - 0 < θ < 0.58745 \dots - + - θ = 0.58745 \dots 0 + 0 0.58745 \dots < θ < \frac{π}{2} + + + θ = \frac{π}{2} + 0 Undefined \frac{π}{2} < θ < π - 0.58745 \dots + - - θ = π - 0.58745 \dots 0 - 0 π - 0.58745 \dots < θ < \frac{3 π}{2} - - + θ = \frac{3 π}{2} - 0 Undefined \frac{3 π}{2} < θ < 2 π - + - θ = 2 π - + -

Identify the intervals that satisfy the required condition:

> 0

0.58745 \dots < θ < \frac{π}{2} or π - 0.58745 \dots < θ < \frac{3 π}{2}

Apply the periodicity of

tan (θ) - \frac{4}{5} cos (θ)

0.58745 \dots + 2 πn < θ < \frac{π}{2} + 2 πn or π - 0.58745 \dots + 2 πn < θ < \frac{3 π}{2} + 2 πn

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