What is the general solution for 1/(1-cos(θ))-1/(1+cos(θ))=2csc^2(θ) ?

The general solution for 1/(1-cos(θ))-1/(1+cos(θ))=2csc^2(θ) is No Solution for θ\in\mathbb{R}

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1/(1-cos(θ))-1/(1+cos(θ))=2csc^2(θ)

Solution

\frac{1}{1 - cos ( θ )} - \frac{1}{1 + cos ( θ )} = 2 csc^{2} (θ)

Solution

No Solution for θ \in R

Solution steps

\frac{1}{1 - cos ( θ )} - \frac{1}{1 + cos ( θ )} = 2 csc^{2} (θ)

Subtract

2 csc^{2} (θ)

from both sides

\frac{2 csc ^{2} ( θ ) cos ^{2} ( θ ) + 2 cos ( θ ) - 2 csc ^{2} ( θ )}{( - cos ( θ ) + 1 ) ( cos ( θ ) + 1 )} = 0

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

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

Rewrite using trig identities

2 cos (θ) - 2 csc^{2} (θ) + 2 cos^{2} (θ) csc^{2} (θ)

Use the Pythagorean identity:

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

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

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

Simplify

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

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

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

Expand

2 csc^{2} (θ) (1 - sin^{2} (θ)) : 2 csc^{2} (θ) - 2 csc^{2} (θ) sin^{2} (θ)

2 csc^{2} (θ) (1 - sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 2 csc^{2} (θ), b = 1, c = sin^{2} (θ)

= 2 csc^{2} (θ) \cdot 1 - 2 csc^{2} (θ) sin^{2} (θ)

= 2 \cdot 1 \cdot csc^{2} (θ) - 2 csc^{2} (θ) sin^{2} (θ)

Multiply the numbers:

2 \cdot 1 = 2

= 2 csc^{2} (θ) - 2 csc^{2} (θ) sin^{2} (θ)

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

Add similar elements:

- 2 csc^{2} (θ) + 2 csc^{2} (θ) = 0

= 2 cos (θ) - 2 csc^{2} (θ) sin^{2} (θ)

= 2 cos (θ) - 2 csc^{2} (θ) sin^{2} (θ)

Use the basic trigonometric identity:

sin (x) = \frac{1}{csc ( x )}

= 2 cos (θ) - 2 csc^{2} (θ) (\frac{1}{csc ( θ )})^{2}

2 csc^{2} (θ) (\frac{1}{csc ( θ )})^{2} = 2

2 csc^{2} (θ) (\frac{1}{csc ( θ )})^{2}

(\frac{1}{csc ( θ )})^{2} = \frac{1}{csc ^{2} ( θ )}

(\frac{1}{csc ( θ )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{csc ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{csc ^{2} ( θ )}

= 2 \cdot \frac{1}{csc ^{2} ( θ )} csc^{2} (θ)

Multiply fractions:

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

= \frac{1 \cdot 2 csc ^{2} ( θ )}{csc ^{2} ( θ )}

Cancel the common factor:

csc^{2} (θ)

= 1 \cdot 2

Multiply the numbers:

1 \cdot 2 = 2

= 2

= 2 cos (θ) - 2

- 2 + 2 cos (θ) = 0

Move

2

to the right side

- 2 + 2 cos (θ) = 0

Add

2

to both sides

- 2 + 2 cos (θ) + 2 = 0 + 2

Simplify

2 cos (θ) = 2

2 cos (θ) = 2

Divide both sides by

2

2 cos (θ) = 2

Divide both sides by

2

\frac{2 cos ( θ )}{2} = \frac{2}{2}

Simplify

cos (θ) = 1

cos (θ) = 1

General solutions for

cos (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}

θ = 0 + 2 πn

θ = 0 + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

Since the equation is undefined for:

2 πn

No Solution for θ \in R

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Frequently Asked Questions (FAQ)

What is the general solution for 1/(1-cos(θ))-1/(1+cos(θ))=2csc^2(θ) ?
The general solution for 1/(1-cos(θ))-1/(1+cos(θ))=2csc^2(θ) is No Solution for θ\in\mathbb{R}