What is the general solution for cot(θ)+2csc(θ)=4 ?

The general solution for cot(θ)+2csc(θ)=4 is θ=0.75142…+2pin,θ=2.88012…+2pin

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Popular Trigonometry >

cot(θ)+2csc(θ)=4

Solution

cot (θ) + 2 csc (θ) = 4

Solution

θ = 0.75142 \dots + 2 πn, θ = 2.88012 \dots + 2 πn

Degrees

θ = 43.05338 \dots^{\circ} + 36 0^{\circ} n, θ = 165.01910 \dots^{\circ} + 36 0^{\circ} n

Solution steps

cot (θ) + 2 csc (θ) = 4

Subtract

4

from both sides

cot (θ) + 2 csc (θ) - 4 = 0

Express with sin, cos

\frac{cos ( θ )}{sin ( θ )} + 2 \cdot \frac{1}{sin ( θ )} - 4 = 0

Simplify

\frac{cos ( θ )}{sin ( θ )} + 2 \cdot \frac{1}{sin ( θ )} - 4 : \frac{cos ( θ ) + 2 - 4 sin ( θ )}{sin ( θ )}

\frac{cos ( θ )}{sin ( θ )} + 2 \cdot \frac{1}{sin ( θ )} - 4

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

2 \cdot \frac{1}{sin ( θ )}

Multiply fractions:

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

= \frac{1 \cdot 2}{sin ( θ )}

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2}{sin ( θ )}

= \frac{cos ( θ )}{sin ( θ )} + \frac{2}{sin ( θ )} - 4

Combine the fractions

\frac{cos ( θ )}{sin ( θ )} + \frac{2}{sin ( θ )} : \frac{cos ( θ ) + 2}{sin ( θ )}

Apply rule

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

= \frac{cos ( θ ) + 2}{sin ( θ )}

= \frac{cos ( θ ) + 2}{sin ( θ )} - 4

Convert element to fraction:

4 = \frac{4 sin ( θ )}{sin ( θ )}

= \frac{cos ( θ ) + 2}{sin ( θ )} - \frac{4 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 ) + 2 - 4 sin ( θ )}{sin ( θ )}

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

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

cos (θ) + 2 - 4 sin (θ) = 0

Add

4 sin (θ)

to both sides

cos (θ) + 2 = 4 sin (θ)

Square both sides

(cos (θ) + 2)^{2} = (4 sin (θ))^{2}

Subtract

(4 sin (θ))^{2}

from both sides

(cos (θ) + 2)^{2} - 16 sin^{2} (θ) = 0

Rewrite using trig identities

(2 + cos (θ))^{2} - 16 sin^{2} (θ)

Use the Pythagorean identity:

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

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

= (2 + cos (θ))^{2} - 16 (1 - cos^{2} (θ))

Simplify

(2 + cos (θ))^{2} - 16 (1 - cos^{2} (θ)) : 17 cos^{2} (θ) + 4 cos (θ) - 12

(2 + cos (θ))^{2} - 16 (1 - cos^{2} (θ))

(2 + cos (θ))^{2} : 4 + 4 cos (θ) + cos^{2} (θ)

Apply Perfect Square Formula:

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

a = 2, b = cos (θ)

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

Simplify

2^{2} + 2 \cdot 2 cos (θ) + cos^{2} (θ) : 4 + 4 cos (θ) + cos^{2} (θ)

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

2^{2} = 4

= 4 + 2 \cdot 2 cos (θ) + cos^{2} (θ)

Multiply the numbers:

2 \cdot 2 = 4

= 4 + 4 cos (θ) + cos^{2} (θ)

= 4 + 4 cos (θ) + cos^{2} (θ)

= 4 + 4 cos (θ) + cos^{2} (θ) - 16 (1 - cos^{2} (θ))

Expand

- 16 (1 - cos^{2} (θ)) : - 16 + 16 cos^{2} (θ)

- 16 (1 - cos^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = - 16, b = 1, c = cos^{2} (θ)

= - 16 \cdot 1 - (- 16) cos^{2} (θ)

Apply minus-plus rules

- (- a) = a

= - 16 \cdot 1 + 16 cos^{2} (θ)

Multiply the numbers:

16 \cdot 1 = 16

= - 16 + 16 cos^{2} (θ)

= 4 + 4 cos (θ) + cos^{2} (θ) - 16 + 16 cos^{2} (θ)

Simplify

4 + 4 cos (θ) + cos^{2} (θ) - 16 + 16 cos^{2} (θ) : 17 cos^{2} (θ) + 4 cos (θ) - 12

4 + 4 cos (θ) + cos^{2} (θ) - 16 + 16 cos^{2} (θ)

Group like terms

= 4 cos (θ) + cos^{2} (θ) + 16 cos^{2} (θ) + 4 - 16

Add similar elements:

cos^{2} (θ) + 16 cos^{2} (θ) = 17 cos^{2} (θ)

= 4 cos (θ) + 17 cos^{2} (θ) + 4 - 16

Add/Subtract the numbers:

4 - 16 = - 12

= 17 cos^{2} (θ) + 4 cos (θ) - 12

= 17 cos^{2} (θ) + 4 cos (θ) - 12

= 17 cos^{2} (θ) + 4 cos (θ) - 12

- 12 + 17 cos^{2} (θ) + 4 cos (θ) = 0

Solve by substitution

- 12 + 17 cos^{2} (θ) + 4 cos (θ) = 0

Let:

cos (θ) = u

- 12 + 17 u^{2} + 4 u = 0

- 12 + 17 u^{2} + 4 u = 0 : u = \frac{2 ( 2 13 - 1 )}{17}, u = - \frac{2 ( 1 + 2 13 )}{17}

- 12 + 17 u^{2} + 4 u = 0

Write in the standard form

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

17 u^{2} + 4 u - 12 = 0

Solve with the quadratic formula

17 u^{2} + 4 u - 12 = 0

Quadratic Equation Formula:

For

a = 17, b = 4, c = - 12

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 17 ( - 12 )}{2 \cdot 17}

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 17 ( - 12 )}{2 \cdot 17}

4^{2} - 4 \cdot 17 (- 12) = 813

4^{2} - 4 \cdot 17 (- 12)

Apply rule

- (- a) = a

= 4^{2} + 4 \cdot 17 \cdot 12

Multiply the numbers:

4 \cdot 17 \cdot 12 = 816

= 4^{2} + 816

4^{2} = 16

= 16 + 816

Add the numbers:

16 + 816 = 832

= 832

Prime factorization of

832 : 2^{6} \cdot 13

832

832

divides by

2832 = 416 \cdot 2

= 2 \cdot 416

416

divides by

2416 = 208 \cdot 2

= 2 \cdot 2 \cdot 208

208

divides by

2208 = 104 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 104

104

divides by

2104 = 52 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 52

52

divides by

252 = 26 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 26

26

divides by

226 = 13 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 13

2, 13

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 13

= 2^{6} \cdot 13

= 2^{6} \cdot 13

Apply radical rule:

n ab = n a n b

= 13 2^{6}

Apply radical rule:

n a^{m} = a^{\frac{m}{n}}

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

= 2^{3} 13

Refine

= 813

u_{1, 2} = \frac{- 4 \pm 8 13}{2 \cdot 17}

Separate the solutions

u_{1} = \frac{- 4 + 8 13}{2 \cdot 17}, u_{2} = \frac{- 4 - 8 13}{2 \cdot 17}

u = \frac{- 4 + 8 13}{2 \cdot 17} : \frac{2 ( 2 13 - 1 )}{17}

\frac{- 4 + 8 13}{2 \cdot 17}

Multiply the numbers:

2 \cdot 17 = 34

= \frac{- 4 + 8 13}{34}

Factor

- 4 + 813 : 4 (- 1 + 213)

- 4 + 813

Rewrite as

= - 4 \cdot 1 + 4 \cdot 213

Factor out common term

4

= 4 (- 1 + 213)

= \frac{4 ( - 1 + 2 13 )}{34}

Cancel the common factor:

2

= \frac{2 ( 2 13 - 1 )}{17}

u = \frac{- 4 - 8 13}{2 \cdot 17} : - \frac{2 ( 1 + 2 13 )}{17}

\frac{- 4 - 8 13}{2 \cdot 17}

Multiply the numbers:

2 \cdot 17 = 34

= \frac{- 4 - 8 13}{34}

Factor

- 4 - 813 : - 4 (1 + 213)

- 4 - 813

Rewrite as

= - 4 \cdot 1 - 4 \cdot 213

Factor out common term

4

= - 4 (1 + 213)

= - \frac{4 ( 1 + 2 13 )}{34}

Cancel the common factor:

2

= - \frac{2 ( 1 + 2 13 )}{17}

The solutions to the quadratic equation are:

u = \frac{2 ( 2 13 - 1 )}{17}, u = - \frac{2 ( 1 + 2 13 )}{17}

Substitute back

u = cos (θ)

cos (θ) = \frac{2 ( 2 13 - 1 )}{17}, cos (θ) = - \frac{2 ( 1 + 2 13 )}{17}

cos (θ) = \frac{2 ( 2 13 - 1 )}{17}, cos (θ) = - \frac{2 ( 1 + 2 13 )}{17}

cos (θ) = \frac{2 ( 2 13 - 1 )}{17} : θ = arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 πn, θ = 2 π - arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 πn

cos (θ) = \frac{2 ( 2 13 - 1 )}{17}

Apply trig inverse properties

cos (θ) = \frac{2 ( 2 13 - 1 )}{17}

General solutions for

cos (θ) = \frac{2 ( 2 13 - 1 )}{17}

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

θ = arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 πn, θ = 2 π - arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 πn

θ = arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 πn, θ = 2 π - arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 πn

cos (θ) = - \frac{2 ( 1 + 2 13 )}{17} : θ = arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 πn, θ = - arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 πn

cos (θ) = - \frac{2 ( 1 + 2 13 )}{17}

Apply trig inverse properties

cos (θ) = - \frac{2 ( 1 + 2 13 )}{17}

General solutions for

cos (θ) = - \frac{2 ( 1 + 2 13 )}{17}

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

θ = arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 πn, θ = - arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 πn

θ = arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 πn, θ = - arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 πn

Combine all the solutions

θ = arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 πn, θ = 2 π - arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 πn, θ = arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 πn, θ = - arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

cot (θ) + 2 csc (θ) = 4

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

Check the solution

arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 πn :

True

arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 πn

Plug in

n = 1

arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 π 1

For

cot (θ) + 2 csc (θ) = 4

plug in

θ = arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 π 1

cot (arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 π 1) + 2 csc (arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 π 1) = 4

Refine

4 = 4

\Rightarrow True

Check the solution

2 π - arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 πn :

False

2 π - arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 πn

Plug in

n = 1

2 π - arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 π 1

For

cot (θ) + 2 csc (θ) = 4

plug in

θ = 2 π - arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 π 1

cot (2 π - arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 π 1) + 2 csc (2 π - arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 π 1) = 4

Refine

- 4 = 4

\Rightarrow False

Check the solution

arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 πn :

True

arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 πn

Plug in

n = 1

arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 π 1

For

cot (θ) + 2 csc (θ) = 4

plug in

θ = arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 π 1

cot (arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 π 1) + 2 csc (arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 π 1) = 4

Refine

4 = 4

\Rightarrow True

Check the solution

- arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 πn :

False

- arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 πn

Plug in

n = 1

- arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 π 1

For

cot (θ) + 2 csc (θ) = 4

plug in

θ = - arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 π 1

cot (- arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 π 1) + 2 csc (- arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 π 1) = 4

Refine

- 4 = 4

\Rightarrow False

θ = arccos (\frac{2 ( 2 13 - 1 )}{17}) + 2 πn, θ = arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 2 πn

Show solutions in decimal form

θ = 0.75142 \dots + 2 πn, θ = 2.88012 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for cot(θ)+2csc(θ)=4 ?
The general solution for cot(θ)+2csc(θ)=4 is θ=0.75142…+2pin,θ=2.88012…+2pin