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Popular >

cos(θ)-5/6 \land tan(θ)<0

Solution

cos (θ) - \frac{5}{6} and tan (θ) < 0

Solution

0.72972 \dots + 2 πn < θ < \frac{π}{2} + 2 πn or π - 0.72972 \dots + 2 πn < θ < \frac{3 π}{2} + 2 πn

Interval Notation

(0.72972 \dots + 2 πn, \frac{π}{2} + 2 πn) \cup (π - 0.72972 \dots + 2 πn, \frac{3 π}{2} + 2 πn)

Decimal Notation

0.72972 \dots + 2 πn < θ < 1.57079 \dots + 2 πn or 2.41186 \dots + 2 πn < θ < 4.71238 \dots + 2 πn

Solution steps

cos (θ) - \frac{5}{6} and tan (θ) < 0

Periodicity of

cos (θ) - \frac{5}{6} and tan (θ) : 2 π

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

cos (θ), \frac{5}{6} and tan (θ)

Periodicity of

cos (θ) : 2 π

Periodicity of

cos (x)

2 π

= 2 π

Periodicity of

\frac{5}{6} and 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:

2 π, π

= 2 π

Express with sin, cos

cos (θ) - \frac{5}{6} and tan (θ) < 0

Use the basic trigonometric identity:

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

cos (θ) - \frac{5}{6} and \frac{sin ( θ )}{cos ( θ )} < 0

cos (θ) - \frac{5}{6} and \frac{sin ( θ )}{cos ( θ )} < 0

Simplify

cos (θ) - \frac{5}{6} and \frac{sin ( θ )}{cos ( θ )} : \frac{6 cos ^{2} ( θ ) - 5 sin ( θ )}{6 cos ( θ )}

cos (θ) - \frac{5}{6} and \frac{sin ( θ )}{cos ( θ )}

Multiply

\frac{5}{6} and \frac{sin ( θ )}{cos ( θ )} : \frac{5 and sin ( θ )}{6 cos ( θ )}

\frac{5}{6} and \frac{sin ( θ )}{cos ( θ )}

Multiply fractions:

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

= \frac{5 sin ( θ ) and}{6 cos ( θ )}

= cos (θ) - \frac{5 sin ( θ ) and}{6 cos ( θ )}

Convert element to fraction:

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

= \frac{cos ( θ ) \cdot 6 cos ( θ )}{6 cos ( θ )} - \frac{5 sin ( θ ) and}{6 cos ( θ )}

Since the denominators are equal, combine the fractions:

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

= \frac{cos ( θ ) \cdot 6 cos ( θ ) - 5 sin ( θ )}{6 cos ( θ )}

cos (θ) \cdot 6 cos (θ) - 5 sin (θ) = 6 cos^{2} (θ) - 5 sin (θ)

cos (θ) \cdot 6 cos (θ) - 5 sin (θ)

cos (θ) \cdot 6 cos (θ) = 6 cos^{2} (θ)

cos (θ) \cdot 6 cos (θ)

Apply exponent rule:

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

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= 6 cos^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= 6 cos^{2} (θ)

= 6 cos^{2} (θ) - 5 sin (θ)

= \frac{6 cos ^{2} ( θ ) - 5 sin ( θ )}{6 cos ( θ )}

\frac{6 cos ^{2} ( θ ) - 5 sin ( θ )}{6 cos ( θ )} < 0

Find the zeroes and undifined points of

\frac{6 cos ^{2} ( θ ) - 5 sin ( θ )}{6 cos ( θ )}

for

0 \leq θ < 2 π

To find the zeroes, set the inequality to zero

\frac{6 cos ^{2} ( θ ) - 5 sin ( θ )}{6 cos ( θ )} = 0

\frac{6 cos ^{2} ( θ ) - 5 sin ( θ )}{6 cos ( θ )} = 0, 0 \leq θ < 2 π : θ = 0.72972 \dots, θ = π - 0.72972 \dots

\frac{6 cos ^{2} ( θ ) - 5 sin ( θ )}{6 cos ( θ )} = 0, 0 \leq θ < 2 π

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

6 cos^{2} (θ) - 5 sin (θ) = 0

Rewrite using trig identities

- 5 sin (θ) + 6 cos^{2} (θ)

Use the Pythagorean identity:

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

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

= - 5 sin (θ) + 6 (1 - sin^{2} (θ))

(1 - sin^{2} (θ)) \cdot 6 - 5 sin (θ) = 0

Solve by substitution

(1 - sin^{2} (θ)) \cdot 6 - 5 sin (θ) = 0

Let:

sin (θ) = u

(1 - u^{2}) \cdot 6 - 5 u = 0

(1 - u^{2}) \cdot 6 - 5 u = 0 : u = - \frac{3}{2}, u = \frac{2}{3}

(1 - u^{2}) \cdot 6 - 5 u = 0

Expand

(1 - u^{2}) \cdot 6 - 5 u : 6 - 6 u^{2} - 5 u

(1 - u^{2}) \cdot 6 - 5 u

= 6 (1 - u^{2}) - 5 u

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}

= 6 - 6 u^{2} - 5 u

6 - 6 u^{2} - 5 u = 0

Write in the standard form

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

- 6 u^{2} - 5 u + 6 = 0

Solve with the quadratic formula

- 6 u^{2} - 5 u + 6 = 0

Quadratic Equation Formula:

For

a = - 6, b = - 5, c = 6

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 6 ) \cdot 6}{2 ( - 6 )}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 6 ) \cdot 6}{2 ( - 6 )}

(- 5)^{2} - 4 (- 6) \cdot 6 = 13

(- 5)^{2} - 4 (- 6) \cdot 6

Apply rule

- (- a) = a

= (- 5)^{2} + 4 \cdot 6 \cdot 6

Apply exponent rule:

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

n

is even

(- 5)^{2} = 5^{2}

= 5^{2} + 4 \cdot 6 \cdot 6

Multiply the numbers:

4 \cdot 6 \cdot 6 = 144

= 5^{2} + 144

5^{2} = 25

= 25 + 144

Add the numbers:

25 + 144 = 169

= 169

Factor the number:

169 = 1 3^{2}

= 1 3^{2}

Apply radical rule:

1 3^{2} = 13

= 13

u_{1, 2} = \frac{- ( - 5 ) \pm 13}{2 ( - 6 )}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 13}{2 ( - 6 )}, u_{2} = \frac{- ( - 5 ) - 13}{2 ( - 6 )}

u = \frac{- ( - 5 ) + 13}{2 ( - 6 )} : - \frac{3}{2}

\frac{- ( - 5 ) + 13}{2 ( - 6 )}

Remove parentheses:

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

= \frac{5 + 13}{- 2 \cdot 6}

Add the numbers:

5 + 13 = 18

= \frac{18}{- 2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{18}{- 12}

Apply the fraction rule:

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

= - \frac{18}{12}

Cancel the common factor:

6

= - \frac{3}{2}

u = \frac{- ( - 5 ) - 13}{2 ( - 6 )} : \frac{2}{3}

\frac{- ( - 5 ) - 13}{2 ( - 6 )}

Remove parentheses:

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

= \frac{5 - 13}{- 2 \cdot 6}

Subtract the numbers:

5 - 13 = - 8

= \frac{- 8}{- 2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 8}{- 12}

Apply the fraction rule:

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

= \frac{8}{12}

Cancel the common factor:

4

= \frac{2}{3}

The solutions to the quadratic equation are:

u = - \frac{3}{2}, u = \frac{2}{3}

Substitute back

u = sin (θ)

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

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

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

No Solution

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

- 1 \leq sin (x) \leq 1

No Solution

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

sin (θ) = \frac{2}{3}, 0 \leq θ < 2 π

Apply trig inverse properties

sin (θ) = \frac{2}{3}

General solutions for

sin (θ) = \frac{2}{3}

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

θ = arcsin (\frac{2}{3}) + 2 πn, θ = π - arcsin (\frac{2}{3}) + 2 πn

θ = arcsin (\frac{2}{3}) + 2 πn, θ = π - arcsin (\frac{2}{3}) + 2 πn

Solutions for the range

0 \leq θ < 2 π

θ = arcsin (\frac{2}{3}), θ = π - arcsin (\frac{2}{3})

Combine all the solutions

θ = arcsin (\frac{2}{3}), θ = π - arcsin (\frac{2}{3})

Show solutions in decimal form

θ = 0.72972 \dots, θ = π - 0.72972 \dots

Find the undefined points:

θ = \frac{π}{2}, θ = \frac{3 π}{2}

Find the zeros of the denominator

6 cos (θ) = 0

Divide both sides by

6

6 cos (θ) = 0

Divide both sides by

6

\frac{6 cos ( θ )}{6} = \frac{0}{6}

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}

0.72972 \dots, \frac{π}{2}, π - 0.72972 \dots, \frac{3 π}{2}

Identify the intervals

0 < θ < 0.72972 \dots, 0.72972 \dots < θ < \frac{π}{2}, \frac{π}{2} < θ < π - 0.72972 \dots, π - 0.72972 \dots < θ < \frac{3 π}{2}, \frac{3 π}{2} < θ < 2 π

Summarize in a table:

6 cos^{2} (θ) - 5 sin (θ) cos (θ) \frac{6 c o s ^{2} ( θ ) - 5 s i n ( θ )}{6 c o s ( θ )} θ = 0 + + + 0 < θ < 0.72972 \dots + + + θ = 0.72972 \dots 0 + 0 0.72972 \dots < θ < \frac{π}{2} - + - θ = \frac{π}{2} - 0 Undefined \frac{π}{2} < θ < π - 0.72972 \dots - - + θ = π - 0.72972 \dots 0 - 0 π - 0.72972 \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

0.72972 \dots < θ < \frac{π}{2} or π - 0.72972 \dots < θ < \frac{3 π}{2}

Apply the periodicity of

cos (θ) - \frac{5}{6} and tan (θ)

0.72972 \dots + 2 πn < θ < \frac{π}{2} + 2 πn or π - 0.72972 \dots + 2 πn < θ < \frac{3 π}{2} + 2 πn

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