What is the general solution for cot(2θ)+csc(2θ)-7=0 ?

The general solution for cot(2θ)+csc(2θ)-7=0 is θ=(0.28379…)/2+pin

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cot(2θ)+csc(2θ)-7=0

Solution

cot (2 θ) + csc (2 θ) - 7 = 0

Solution

θ = \frac{0.28379 \dots}{2} + πn

Degrees

θ = 8.13010 \dots^{\circ} + 18 0^{\circ} n

Solution steps

cot (2 θ) + csc (2 θ) - 7 = 0

Express with sin, cos

\frac{cos ( 2 θ )}{sin ( 2 θ )} + \frac{1}{sin ( 2 θ )} - 7 = 0

Simplify

\frac{cos ( 2 θ )}{sin ( 2 θ )} + \frac{1}{sin ( 2 θ )} - 7 : \frac{cos ( 2 θ ) + 1 - 7 sin ( 2 θ )}{sin ( 2 θ )}

\frac{cos ( 2 θ )}{sin ( 2 θ )} + \frac{1}{sin ( 2 θ )} - 7

Combine the fractions

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

Apply rule

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

= \frac{cos ( 2 θ ) + 1}{sin ( 2 θ )}

= \frac{cos ( 2 θ ) + 1}{sin ( 2 θ )} - 7

Convert element to fraction:

7 = \frac{7 sin ( 2 θ )}{sin ( 2 θ )}

= \frac{cos ( 2 θ ) + 1}{sin ( 2 θ )} - \frac{7 sin ( 2 θ )}{sin ( 2 θ )}

Since the denominators are equal, combine the fractions:

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

= \frac{cos ( 2 θ ) + 1 - 7 sin ( 2 θ )}{sin ( 2 θ )}

\frac{cos ( 2 θ ) + 1 - 7 sin ( 2 θ )}{sin ( 2 θ )} = 0

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

cos (2 θ) + 1 - 7 sin (2 θ) = 0

Add

7 sin (2 θ)

to both sides

cos (2 θ) + 1 = 7 sin (2 θ)

Square both sides

(cos (2 θ) + 1)^{2} = (7 sin (2 θ))^{2}

Subtract

(7 sin (2 θ))^{2}

from both sides

(cos (2 θ) + 1)^{2} - 49 sin^{2} (2 θ) = 0

Rewrite using trig identities

(1 + cos (2 θ))^{2} - 49 sin^{2} (2 θ)

Use the Pythagorean identity:

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

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

= (1 + cos (2 θ))^{2} - 49 (1 - cos^{2} (2 θ))

Simplify

(1 + cos (2 θ))^{2} - 49 (1 - cos^{2} (2 θ)) : 50 cos^{2} (2 θ) + 2 cos (2 θ) - 48

(1 + cos (2 θ))^{2} - 49 (1 - cos^{2} (2 θ))

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

Apply Perfect Square Formula:

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

a = 1, b = cos (2 θ)

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

Simplify

1^{2} + 2 \cdot 1 \cdot cos (2 θ) + cos^{2} (2 θ) : 1 + 2 cos (2 θ) + cos^{2} (2 θ)

1^{2} + 2 \cdot 1 \cdot cos (2 θ) + cos^{2} (2 θ)

Apply rule

1^{a} = 1

1^{2} = 1

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

Multiply the numbers:

2 \cdot 1 = 2

= 1 + 2 cos (2 θ) + cos^{2} (2 θ)

= 1 + 2 cos (2 θ) + cos^{2} (2 θ)

= 1 + 2 cos (2 θ) + cos^{2} (2 θ) - 49 (1 - cos^{2} (2 θ))

Expand

- 49 (1 - cos^{2} (2 θ)) : - 49 + 49 cos^{2} (2 θ)

- 49 (1 - cos^{2} (2 θ))

Apply the distributive law:

a (b - c) = ab - a c

a = - 49, b = 1, c = cos^{2} (2 θ)

= - 49 \cdot 1 - (- 49) cos^{2} (2 θ)

Apply minus-plus rules

- (- a) = a

= - 49 \cdot 1 + 49 cos^{2} (2 θ)

Multiply the numbers:

49 \cdot 1 = 49

= - 49 + 49 cos^{2} (2 θ)

= 1 + 2 cos (2 θ) + cos^{2} (2 θ) - 49 + 49 cos^{2} (2 θ)

Simplify

1 + 2 cos (2 θ) + cos^{2} (2 θ) - 49 + 49 cos^{2} (2 θ) : 50 cos^{2} (2 θ) + 2 cos (2 θ) - 48

1 + 2 cos (2 θ) + cos^{2} (2 θ) - 49 + 49 cos^{2} (2 θ)

Group like terms

= 2 cos (2 θ) + cos^{2} (2 θ) + 49 cos^{2} (2 θ) + 1 - 49

Add similar elements:

cos^{2} (2 θ) + 49 cos^{2} (2 θ) = 50 cos^{2} (2 θ)

= 2 cos (2 θ) + 50 cos^{2} (2 θ) + 1 - 49

Add/Subtract the numbers:

1 - 49 = - 48

= 50 cos^{2} (2 θ) + 2 cos (2 θ) - 48

= 50 cos^{2} (2 θ) + 2 cos (2 θ) - 48

= 50 cos^{2} (2 θ) + 2 cos (2 θ) - 48

- 48 + 2 cos (2 θ) + 50 cos^{2} (2 θ) = 0

Solve by substitution

- 48 + 2 cos (2 θ) + 50 cos^{2} (2 θ) = 0

Let:

cos (2 θ) = u

- 48 + 2 u + 50 u^{2} = 0

- 48 + 2 u + 50 u^{2} = 0 : u = \frac{24}{25}, u = - 1

- 48 + 2 u + 50 u^{2} = 0

Write in the standard form

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

50 u^{2} + 2 u - 48 = 0

Solve with the quadratic formula

50 u^{2} + 2 u - 48 = 0

Quadratic Equation Formula:

For

a = 50, b = 2, c = - 48

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 50 ( - 48 )}{2 \cdot 50}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 50 ( - 48 )}{2 \cdot 50}

2^{2} - 4 \cdot 50 (- 48) = 98

2^{2} - 4 \cdot 50 (- 48)

Apply rule

- (- a) = a

= 2^{2} + 4 \cdot 50 \cdot 48

Multiply the numbers:

4 \cdot 50 \cdot 48 = 9600

= 2^{2} + 9600

2^{2} = 4

= 4 + 9600

Add the numbers:

4 + 9600 = 9604

= 9604

Factor the number:

9604 = 9 8^{2}

= 9 8^{2}

Apply radical rule:

9 8^{2} = 98

= 98

u_{1, 2} = \frac{- 2 \pm 98}{2 \cdot 50}

Separate the solutions

u_{1} = \frac{- 2 + 98}{2 \cdot 50}, u_{2} = \frac{- 2 - 98}{2 \cdot 50}

u = \frac{- 2 + 98}{2 \cdot 50} : \frac{24}{25}

\frac{- 2 + 98}{2 \cdot 50}

Add/Subtract the numbers:

- 2 + 98 = 96

= \frac{96}{2 \cdot 50}

Multiply the numbers:

2 \cdot 50 = 100

= \frac{96}{100}

Cancel the common factor:

4

= \frac{24}{25}

u = \frac{- 2 - 98}{2 \cdot 50} : - 1

\frac{- 2 - 98}{2 \cdot 50}

Subtract the numbers:

- 2 - 98 = - 100

= \frac{- 100}{2 \cdot 50}

Multiply the numbers:

2 \cdot 50 = 100

= \frac{- 100}{100}

Apply the fraction rule:

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

= - \frac{100}{100}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = \frac{24}{25}, u = - 1

Substitute back

u = cos (2 θ)

cos (2 θ) = \frac{24}{25}, cos (2 θ) = - 1

cos (2 θ) = \frac{24}{25}, cos (2 θ) = - 1

cos (2 θ) = \frac{24}{25} : θ = \frac{arccos ( \frac{24}{25} )}{2} + πn, θ = π - \frac{arccos ( \frac{24}{25} )}{2} + πn

cos (2 θ) = \frac{24}{25}

Apply trig inverse properties

cos (2 θ) = \frac{24}{25}

General solutions for

cos (2 θ) = \frac{24}{25}

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

2 θ = arccos (\frac{24}{25}) + 2 πn, 2 θ = 2 π - arccos (\frac{24}{25}) + 2 πn

2 θ = arccos (\frac{24}{25}) + 2 πn, 2 θ = 2 π - arccos (\frac{24}{25}) + 2 πn

Solve

2 θ = arccos (\frac{24}{25}) + 2 πn : θ = \frac{arccos ( \frac{24}{25} )}{2} + πn

2 θ = arccos (\frac{24}{25}) + 2 πn

Divide both sides by

2

2 θ = arccos (\frac{24}{25}) + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{arccos ( \frac{24}{25} )}{2} + \frac{2 πn}{2}

Simplify

θ = \frac{arccos ( \frac{24}{25} )}{2} + πn

θ = \frac{arccos ( \frac{24}{25} )}{2} + πn

Solve

2 θ = 2 π - arccos (\frac{24}{25}) + 2 πn : θ = π - \frac{arccos ( \frac{24}{25} )}{2} + πn

2 θ = 2 π - arccos (\frac{24}{25}) + 2 πn

Divide both sides by

2

2 θ = 2 π - arccos (\frac{24}{25}) + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{2 π}{2} - \frac{arccos ( \frac{24}{25} )}{2} + \frac{2 πn}{2}

Simplify

θ = π - \frac{arccos ( \frac{24}{25} )}{2} + πn

θ = π - \frac{arccos ( \frac{24}{25} )}{2} + πn

θ = \frac{arccos ( \frac{24}{25} )}{2} + πn, θ = π - \frac{arccos ( \frac{24}{25} )}{2} + πn

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

cos (2 θ) = - 1

General solutions for

cos (2 θ) = - 1

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}

2 θ = π + 2 πn

2 θ = π + 2 πn

Solve

2 θ = π + 2 πn : θ = \frac{π}{2} + πn

2 θ = π + 2 πn

Divide both sides by

2

2 θ = π + 2 πn

Divide both sides by

2

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

Simplify

θ = \frac{π}{2} + πn

θ = \frac{π}{2} + πn

θ = \frac{π}{2} + πn

Combine all the solutions

θ = \frac{arccos ( \frac{24}{25} )}{2} + πn, θ = π - \frac{arccos ( \frac{24}{25} )}{2} + πn, θ = \frac{π}{2} + πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

cot (2 θ) + csc (2 θ) - 7 = 0

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

Check the solution

\frac{arccos ( \frac{24}{25} )}{2} + πn :

True

\frac{arccos ( \frac{24}{25} )}{2} + πn

Plug in

n = 1

\frac{arccos ( \frac{24}{25} )}{2} + π 1

For

cot (2 θ) + csc (2 θ) - 7 = 0

plug in

θ = \frac{arccos ( \frac{24}{25} )}{2} + π 1

cot (2 (\frac{arccos ( \frac{24}{25} )}{2} + π 1)) + csc (2 (\frac{arccos ( \frac{24}{25} )}{2} + π 1)) - 7 = 0

Refine

0 = 0

\Rightarrow True

Check the solution

π - \frac{arccos ( \frac{24}{25} )}{2} + πn :

False

π - \frac{arccos ( \frac{24}{25} )}{2} + πn

Plug in

n = 1

π - \frac{arccos ( \frac{24}{25} )}{2} + π 1

For

cot (2 θ) + csc (2 θ) - 7 = 0

plug in

θ = π - \frac{arccos ( \frac{24}{25} )}{2} + π 1

cot (2 (π - \frac{arccos ( \frac{24}{25} )}{2} + π 1)) + csc (2 (π - \frac{arccos ( \frac{24}{25} )}{2} + π 1)) - 7 = 0

Refine

- 14 = 0

\Rightarrow False

Check the solution

\frac{π}{2} + πn :

False

\frac{π}{2} + πn

Plug in

n = 1

\frac{π}{2} + π 1

For

cot (2 θ) + csc (2 θ) - 7 = 0

plug in

θ = \frac{π}{2} + π 1

cot (2 (\frac{π}{2} + π 1)) + csc (2 (\frac{π}{2} + π 1)) - 7 = 0

Undefined

\Rightarrow False

θ = \frac{arccos ( \frac{24}{25} )}{2} + πn

Show solutions in decimal form

θ = \frac{0.28379 \dots}{2} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for cot(2θ)+csc(2θ)-7=0 ?
The general solution for cot(2θ)+csc(2θ)-7=0 is θ=(0.28379…)/2+pin