What is the general solution for cot(θ)+5csc(θ)=6 ?

The general solution for cot(θ)+5csc(θ)=6 is θ=1.13005…+2pin,θ=2.34183…+2pin

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Popular Trigonometry >

cot(θ)+5csc(θ)=6

Solution

cot (θ) + 5 csc (θ) = 6

Solution

θ = 1.13005 \dots + 2 πn, θ = 2.34183 \dots + 2 πn

Degrees

θ = 64.74731 \dots^{\circ} + 36 0^{\circ} n, θ = 134.17732 \dots^{\circ} + 36 0^{\circ} n

Solution steps

cot (θ) + 5 csc (θ) = 6

Subtract

6

from both sides

cot (θ) + 5 csc (θ) - 6 = 0

Express with sin, cos

\frac{cos ( θ )}{sin ( θ )} + 5 \cdot \frac{1}{sin ( θ )} - 6 = 0

Simplify

\frac{cos ( θ )}{sin ( θ )} + 5 \cdot \frac{1}{sin ( θ )} - 6 : \frac{cos ( θ ) + 5 - 6 sin ( θ )}{sin ( θ )}

\frac{cos ( θ )}{sin ( θ )} + 5 \cdot \frac{1}{sin ( θ )} - 6

5 \cdot \frac{1}{sin ( θ )} = \frac{5}{sin ( θ )}

5 \cdot \frac{1}{sin ( θ )}

Multiply fractions:

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

= \frac{1 \cdot 5}{sin ( θ )}

Multiply the numbers:

1 \cdot 5 = 5

= \frac{5}{sin ( θ )}

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

Combine the fractions

\frac{cos ( θ )}{sin ( θ )} + \frac{5}{sin ( θ )} : \frac{cos ( θ ) + 5}{sin ( θ )}

Apply rule

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

= \frac{cos ( θ ) + 5}{sin ( θ )}

= \frac{cos ( θ ) + 5}{sin ( θ )} - 6

Convert element to fraction:

6 = \frac{6 sin ( θ )}{sin ( θ )}

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

Since the denominators are equal, combine the fractions:

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

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

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

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

cos (θ) + 5 - 6 sin (θ) = 0

Add

6 sin (θ)

to both sides

cos (θ) + 5 = 6 sin (θ)

Square both sides

(cos (θ) + 5)^{2} = (6 sin (θ))^{2}

Subtract

(6 sin (θ))^{2}

from both sides

(cos (θ) + 5)^{2} - 36 sin^{2} (θ) = 0

Rewrite using trig identities

(5 + cos (θ))^{2} - 36 sin^{2} (θ)

Use the Pythagorean identity:

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

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

= (5 + cos (θ))^{2} - 36 (1 - cos^{2} (θ))

Simplify

(5 + cos (θ))^{2} - 36 (1 - cos^{2} (θ)) : 37 cos^{2} (θ) + 10 cos (θ) - 11

(5 + cos (θ))^{2} - 36 (1 - cos^{2} (θ))

(5 + cos (θ))^{2} : 25 + 10 cos (θ) + cos^{2} (θ)

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = 5, b = cos (θ)

= 5^{2} + 2 \cdot 5 cos (θ) + cos^{2} (θ)

Simplify

5^{2} + 2 \cdot 5 cos (θ) + cos^{2} (θ) : 25 + 10 cos (θ) + cos^{2} (θ)

5^{2} + 2 \cdot 5 cos (θ) + cos^{2} (θ)

5^{2} = 25

= 25 + 2 \cdot 5 cos (θ) + cos^{2} (θ)

Multiply the numbers:

2 \cdot 5 = 10

= 25 + 10 cos (θ) + cos^{2} (θ)

= 25 + 10 cos (θ) + cos^{2} (θ)

= 25 + 10 cos (θ) + cos^{2} (θ) - 36 (1 - cos^{2} (θ))

Expand

- 36 (1 - cos^{2} (θ)) : - 36 + 36 cos^{2} (θ)

- 36 (1 - cos^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = - 36, b = 1, c = cos^{2} (θ)

= - 36 \cdot 1 - (- 36) cos^{2} (θ)

Apply minus-plus rules

- (- a) = a

= - 36 \cdot 1 + 36 cos^{2} (θ)

Multiply the numbers:

36 \cdot 1 = 36

= - 36 + 36 cos^{2} (θ)

= 25 + 10 cos (θ) + cos^{2} (θ) - 36 + 36 cos^{2} (θ)

Simplify

25 + 10 cos (θ) + cos^{2} (θ) - 36 + 36 cos^{2} (θ) : 37 cos^{2} (θ) + 10 cos (θ) - 11

25 + 10 cos (θ) + cos^{2} (θ) - 36 + 36 cos^{2} (θ)

Group like terms

= 10 cos (θ) + cos^{2} (θ) + 36 cos^{2} (θ) + 25 - 36

Add similar elements:

cos^{2} (θ) + 36 cos^{2} (θ) = 37 cos^{2} (θ)

= 10 cos (θ) + 37 cos^{2} (θ) + 25 - 36

Add/Subtract the numbers:

25 - 36 = - 11

= 37 cos^{2} (θ) + 10 cos (θ) - 11

= 37 cos^{2} (θ) + 10 cos (θ) - 11

= 37 cos^{2} (θ) + 10 cos (θ) - 11

- 11 + 10 cos (θ) + 37 cos^{2} (θ) = 0

Solve by substitution

- 11 + 10 cos (θ) + 37 cos^{2} (θ) = 0

Let:

cos (θ) = u

- 11 + 10 u + 37 u^{2} = 0

- 11 + 10 u + 37 u^{2} = 0 : u = \frac{- 5 + 12 3}{37}, u = - \frac{5 + 12 3}{37}

- 11 + 10 u + 37 u^{2} = 0

Write in the standard form

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

37 u^{2} + 10 u - 11 = 0

Solve with the quadratic formula

37 u^{2} + 10 u - 11 = 0

Quadratic Equation Formula:

For

a = 37, b = 10, c = - 11

u_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 37 ( - 11 )}{2 \cdot 37}

u_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 37 ( - 11 )}{2 \cdot 37}

1 0^{2} - 4 \cdot 37 (- 11) = 243

1 0^{2} - 4 \cdot 37 (- 11)

Apply rule

- (- a) = a

= 1 0^{2} + 4 \cdot 37 \cdot 11

Multiply the numbers:

4 \cdot 37 \cdot 11 = 1628

= 1 0^{2} + 1628

1 0^{2} = 100

= 100 + 1628

Add the numbers:

100 + 1628 = 1728

= 1728

Prime factorization of

1728 : 2^{6} \cdot 3^{3}

1728

1728

divides by

21728 = 864 \cdot 2

= 2 \cdot 864

864

divides by

2864 = 432 \cdot 2

= 2 \cdot 2 \cdot 432

432

divides by

2432 = 216 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 216

216

divides by

2216 = 108 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 108

108

divides by

2108 = 54 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 54

54

divides by

254 = 27 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 27

27

divides by

327 = 9 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 9

9

divides by

39 = 3 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3

= 2^{6} \cdot 3^{3}

= 2^{6} \cdot 3^{3}

Apply exponent rule:

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

= 2^{6} \cdot 3^{2} \cdot 3

Apply radical rule:

= 3 2^{6} 3^{2}

Apply radical rule:

2^{6} = 2^{\frac{6}{2}} = 2^{3}

= 2^{3} 3 3^{2}

Apply radical rule:

3^{2} = 3

= 2^{3} \cdot 33

Refine

= 243

u_{1, 2} = \frac{- 10 \pm 24 3}{2 \cdot 37}

Separate the solutions

u_{1} = \frac{- 10 + 24 3}{2 \cdot 37}, u_{2} = \frac{- 10 - 24 3}{2 \cdot 37}

u = \frac{- 10 + 24 3}{2 \cdot 37} : \frac{- 5 + 12 3}{37}

\frac{- 10 + 24 3}{2 \cdot 37}

Multiply the numbers:

2 \cdot 37 = 74

= \frac{- 10 + 24 3}{74}

Factor

- 10 + 243 : 2 (- 5 + 123)

- 10 + 243

Rewrite as

= - 2 \cdot 5 + 2 \cdot 123

Factor out common term

2

= 2 (- 5 + 123)

= \frac{2 ( - 5 + 12 3 )}{74}

Cancel the common factor:

2

= \frac{- 5 + 12 3}{37}

u = \frac{- 10 - 24 3}{2 \cdot 37} : - \frac{5 + 12 3}{37}

\frac{- 10 - 24 3}{2 \cdot 37}

Multiply the numbers:

2 \cdot 37 = 74

= \frac{- 10 - 24 3}{74}

Factor

- 10 - 243 : - 2 (5 + 123)

- 10 - 243

Rewrite as

= - 2 \cdot 5 - 2 \cdot 123

Factor out common term

2

= - 2 (5 + 123)

= - \frac{2 ( 5 + 12 3 )}{74}

Cancel the common factor:

2

= - \frac{5 + 12 3}{37}

The solutions to the quadratic equation are:

u = \frac{- 5 + 12 3}{37}, u = - \frac{5 + 12 3}{37}

Substitute back

u = cos (θ)

cos (θ) = \frac{- 5 + 12 3}{37}, cos (θ) = - \frac{5 + 12 3}{37}

cos (θ) = \frac{- 5 + 12 3}{37}, cos (θ) = - \frac{5 + 12 3}{37}

cos (θ) = \frac{- 5 + 12 3}{37} : θ = arccos (\frac{- 5 + 12 3}{37}) + 2 πn, θ = 2 π - arccos (\frac{- 5 + 12 3}{37}) + 2 πn

cos (θ) = \frac{- 5 + 12 3}{37}

Apply trig inverse properties

cos (θ) = \frac{- 5 + 12 3}{37}

General solutions for

cos (θ) = \frac{- 5 + 12 3}{37}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

θ = arccos (\frac{- 5 + 12 3}{37}) + 2 πn, θ = 2 π - arccos (\frac{- 5 + 12 3}{37}) + 2 πn

θ = arccos (\frac{- 5 + 12 3}{37}) + 2 πn, θ = 2 π - arccos (\frac{- 5 + 12 3}{37}) + 2 πn

cos (θ) = - \frac{5 + 12 3}{37} : θ = arccos (- \frac{5 + 12 3}{37}) + 2 πn, θ = - arccos (- \frac{5 + 12 3}{37}) + 2 πn

cos (θ) = - \frac{5 + 12 3}{37}

Apply trig inverse properties

cos (θ) = - \frac{5 + 12 3}{37}

General solutions for

cos (θ) = - \frac{5 + 12 3}{37}

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

θ = arccos (- \frac{5 + 12 3}{37}) + 2 πn, θ = - arccos (- \frac{5 + 12 3}{37}) + 2 πn

θ = arccos (- \frac{5 + 12 3}{37}) + 2 πn, θ = - arccos (- \frac{5 + 12 3}{37}) + 2 πn

Combine all the solutions

θ = arccos (\frac{- 5 + 12 3}{37}) + 2 πn, θ = 2 π - arccos (\frac{- 5 + 12 3}{37}) + 2 πn, θ = arccos (- \frac{5 + 12 3}{37}) + 2 πn, θ = - arccos (- \frac{5 + 12 3}{37}) + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

cot (θ) + 5 csc (θ) = 6

Remove the ones that don't agree with the equation.

Check the solution

arccos (\frac{- 5 + 12 3}{37}) + 2 πn :

True

arccos (\frac{- 5 + 12 3}{37}) + 2 πn

Plug in

n = 1

arccos (\frac{- 5 + 12 3}{37}) + 2 π 1

For

cot (θ) + 5 csc (θ) = 6

plug in

θ = arccos (\frac{- 5 + 12 3}{37}) + 2 π 1

cot (arccos (\frac{- 5 + 12 3}{37}) + 2 π 1) + 5 csc (arccos (\frac{- 5 + 12 3}{37}) + 2 π 1) = 6

Refine

6 = 6

\Rightarrow True

Check the solution

2 π - arccos (\frac{- 5 + 12 3}{37}) + 2 πn :

False

2 π - arccos (\frac{- 5 + 12 3}{37}) + 2 πn

Plug in

n = 1

2 π - arccos (\frac{- 5 + 12 3}{37}) + 2 π 1

For

cot (θ) + 5 csc (θ) = 6

plug in

θ = 2 π - arccos (\frac{- 5 + 12 3}{37}) + 2 π 1

cot (2 π - arccos (\frac{- 5 + 12 3}{37}) + 2 π 1) + 5 csc (2 π - arccos (\frac{- 5 + 12 3}{37}) + 2 π 1) = 6

Refine

- 6 = 6

\Rightarrow False

Check the solution

arccos (- \frac{5 + 12 3}{37}) + 2 πn :

True

arccos (- \frac{5 + 12 3}{37}) + 2 πn

Plug in

n = 1

arccos (- \frac{5 + 12 3}{37}) + 2 π 1

For

cot (θ) + 5 csc (θ) = 6

plug in

θ = arccos (- \frac{5 + 12 3}{37}) + 2 π 1

cot (arccos (- \frac{5 + 12 3}{37}) + 2 π 1) + 5 csc (arccos (- \frac{5 + 12 3}{37}) + 2 π 1) = 6

Refine

6 = 6

\Rightarrow True

Check the solution

- arccos (- \frac{5 + 12 3}{37}) + 2 πn :

False

- arccos (- \frac{5 + 12 3}{37}) + 2 πn

Plug in

n = 1

- arccos (- \frac{5 + 12 3}{37}) + 2 π 1

For

cot (θ) + 5 csc (θ) = 6

plug in

θ = - arccos (- \frac{5 + 12 3}{37}) + 2 π 1

cot (- arccos (- \frac{5 + 12 3}{37}) + 2 π 1) + 5 csc (- arccos (- \frac{5 + 12 3}{37}) + 2 π 1) = 6

Refine

- 6 = 6

\Rightarrow False

θ = arccos (\frac{- 5 + 12 3}{37}) + 2 πn, θ = arccos (- \frac{5 + 12 3}{37}) + 2 πn

Show solutions in decimal form

θ = 1.13005 \dots + 2 πn, θ = 2.34183 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for cot(θ)+5csc(θ)=6 ?
The general solution for cot(θ)+5csc(θ)=6 is θ=1.13005…+2pin,θ=2.34183…+2pin