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

The general solution for 1/(1-cos(x))+1/(1+cos(x))=2csc(x) is x= pi/2+2pin

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1/(1-cos(x))+1/(1+cos(x))=2csc(x)

Solution

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

Solution

x = \frac{π}{2} + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

2 csc (x)

from both sides

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

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

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

Express with sin, cos

2 cos^{2} (x) \frac{1}{sin ( x )} - 2 \cdot \frac{1}{sin ( x )} + 2 = 0

Simplify

2 cos^{2} (x) \frac{1}{sin ( x )} - 2 \cdot \frac{1}{sin ( x )} + 2 : \frac{2 cos ^{2} ( x ) - 2 + 2 sin ( x )}{sin ( x )}

2 cos^{2} (x) \frac{1}{sin ( x )} - 2 \cdot \frac{1}{sin ( x )} + 2

2 cos^{2} (x) \frac{1}{sin ( x )} = \frac{2 cos ^{2} ( x )}{sin ( x )}

2 cos^{2} (x) \frac{1}{sin ( x )}

Multiply fractions:

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

= \frac{1 \cdot 2 cos ^{2} ( x )}{sin ( x )}

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2 cos ^{2} ( x )}{sin ( x )}

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

2 \cdot \frac{1}{sin ( x )}

Multiply fractions:

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

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

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2}{sin ( x )}

= \frac{2 cos ^{2} ( x )}{sin ( x )} - \frac{2}{sin ( x )} + 2

Combine the fractions

\frac{2 cos ^{2} ( x )}{sin ( x )} - \frac{2}{sin ( x )} : \frac{2 cos ^{2} ( x ) - 2}{sin ( x )}

Apply rule

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

= \frac{2 cos ^{2} ( x ) - 2}{sin ( x )}

= \frac{2 cos ^{2} ( x ) - 2}{sin ( x )} + 2

Convert element to fraction:

2 = \frac{2 sin ( x )}{sin ( x )}

= \frac{2 cos ^{2} ( x ) - 2}{sin ( x )} + \frac{2 sin ( x )}{sin ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{2 cos ^{2} ( x ) - 2 + 2 sin ( x )}{sin ( x )}

\frac{2 cos ^{2} ( x ) - 2 + 2 sin ( x )}{sin ( x )} = 0

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

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

Subtract

2 sin (x)

from both sides

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

Square both sides

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

Subtract

(- 2 sin (x))^{2}

from both sides

(2 cos^{2} (x) - 2)^{2} - 4 sin^{2} (x) = 0

Factor

(2 cos^{2} (x) - 2)^{2} - 4 sin^{2} (x) : 4 (cos^{2} (x) - 1 + sin (x)) (cos^{2} (x) - 1 - sin (x))

(2 cos^{2} (x) - 2)^{2} - 4 sin^{2} (x)

Rewrite

(2 cos^{2} (x) - 2)^{2} - 4 sin^{2} (x)

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

(2 cos^{2} (x) - 2)^{2} - 4 sin^{2} (x)

Rewrite

4

2^{2}

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

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

2^{2} sin^{2} (x) = (2 sin (x))^{2}

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

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

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

Refine

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

Factor

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

2 cos^{2} (x) - 2 + 2 sin (x)

Factor out common term

2

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

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

Factor

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

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

Factor out common term

2

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

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

Refine

= 4 (cos^{2} (x) - 1 + sin (x)) (cos^{2} (x) - 1 - sin (x))

4 (cos^{2} (x) - 1 + sin (x)) (cos^{2} (x) - 1 - sin (x)) = 0

Solving each part separately

cos^{2} (x) - 1 + sin (x) = 0 or cos^{2} (x) - 1 - sin (x) = 0

cos^{2} (x) - 1 + sin (x) = 0 : x = 2 πn, x = π + 2 πn, x = \frac{π}{2} + 2 πn

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

sin (x) - sin^{2} (x) = 0

Solve by substitution

sin (x) - sin^{2} (x) = 0

Let:

sin (x) = u

u - u^{2} = 0

u - u^{2} = 0 : u = 0, u = 1

u - u^{2} = 0

Write in the standard form

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

- u^{2} + u = 0

Solve with the quadratic formula

- u^{2} + u = 0

Quadratic Equation Formula:

For

a = - 1, b = 1, c = 0

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

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

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

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) \cdot 0

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 1 + 0

Add the numbers:

1 + 0 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

u = \frac{- 1 + 1}{2 ( - 1 )} : 0

\frac{- 1 + 1}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 1}{- 2 \cdot 1}

Add/Subtract the numbers:

- 1 + 1 = 0

= \frac{0}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{- 2}

Apply the fraction rule:

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

= - \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

u = \frac{- 1 - 1}{2 ( - 1 )} : 1

\frac{- 1 - 1}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 1}{- 2 \cdot 1}

Subtract the numbers:

- 1 - 1 = - 2

= \frac{- 2}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{- 2}

Apply the fraction rule:

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

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = 0, u = 1

Substitute back

u = sin (x)

sin (x) = 0, sin (x) = 1

sin (x) = 0, sin (x) = 1

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (x) = 0

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

General solutions for

sin (x) = 1

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = \frac{π}{2} + 2 πn

cos^{2} (x) - 1 - sin (x) = 0 : x = \frac{3 π}{2} + 2 πn, x = 2 πn, x = π + 2 πn

cos^{2} (x) - 1 - sin (x) = 0

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

- sin (x) - sin^{2} (x) = 0

Solve by substitution

- sin (x) - sin^{2} (x) = 0

Let:

sin (x) = u

- u - u^{2} = 0

- u - u^{2} = 0 : u = - 1, u = 0

- u - u^{2} = 0

Write in the standard form

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

- u^{2} - u = 0

Solve with the quadratic formula

- u^{2} - u = 0

Quadratic Equation Formula:

For

a = - 1, b = - 1, c = 0

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

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

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

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

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 1 \cdot 0

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 1 \cdot 0 = 0

4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 0

= 1 + 0

Add the numbers:

1 + 0 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

u = \frac{- ( - 1 ) + 1}{2 ( - 1 )} : - 1

\frac{- ( - 1 ) + 1}{2 ( - 1 )}

Remove parentheses:

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

= \frac{1 + 1}{- 2 \cdot 1}

Add the numbers:

1 + 1 = 2

= \frac{2}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{- 2}

Apply the fraction rule:

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

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 1 ) - 1}{2 ( - 1 )} : 0

\frac{- ( - 1 ) - 1}{2 ( - 1 )}

Remove parentheses:

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

= \frac{1 - 1}{- 2 \cdot 1}

Subtract the numbers:

1 - 1 = 0

= \frac{0}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{- 2}

Apply the fraction rule:

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

= - \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

The solutions to the quadratic equation are:

u = - 1, u = 0

Substitute back

u = sin (x)

sin (x) = - 1, sin (x) = 0

sin (x) = - 1, sin (x) = 0

sin (x) = - 1 : x = \frac{3 π}{2} + 2 πn

sin (x) = - 1

General solutions for

sin (x) = - 1

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (x) = 0

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

Combine all the solutions

x = \frac{3 π}{2} + 2 πn, x = 2 πn, x = π + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

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

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

Check the solution

2 πn :

False

2 πn

Plug in

n = 1

2 π 1

For

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

plug in

x = 2 π 1

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

Undefined

\Rightarrow False

Check the solution

π + 2 πn :

False

π + 2 πn

Plug in

n = 1

π + 2 π 1

For

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

plug in

x = π + 2 π 1

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

Undefined

\Rightarrow False

Check the solution

\frac{π}{2} + 2 πn :

True

\frac{π}{2} + 2 πn

Plug in

n = 1

\frac{π}{2} + 2 π 1

For

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

plug in

x = \frac{π}{2} + 2 π 1

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

Refine

2 = 2

\Rightarrow True

Check the solution

\frac{3 π}{2} + 2 πn :

False

\frac{3 π}{2} + 2 πn

Plug in

n = 1

\frac{3 π}{2} + 2 π 1

For

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

plug in

x = \frac{3 π}{2} + 2 π 1

\frac{1}{1 - cos ( \frac{3 π}{2} + 2 π 1 )} + \frac{1}{1 + cos ( \frac{3 π}{2} + 2 π 1 )} = 2 csc (\frac{3 π}{2} + 2 π 1)

Refine

2 = - 2

\Rightarrow False

x = \frac{π}{2} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 1/(1-cos(x))+1/(1+cos(x))=2csc(x) ?
The general solution for 1/(1-cos(x))+1/(1+cos(x))=2csc(x) is x= pi/2+2pin