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

The general solution for csc(2χ)= 1/(1-cos(2χ)) is χ= pi/4+pin

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csc(2χ)= 1/(1-cos(2χ))

Solution

csc (2 χ) = \frac{1}{1 - cos ( 2 χ )}

Solution

χ = \frac{π}{4} + πn

Degrees

χ = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

csc (2 χ) = \frac{1}{1 - cos ( 2 χ )}

Subtract

\frac{1}{1 - cos ( 2 χ )}

from both sides

csc (2 χ) - \frac{1}{1 - cos ( 2 χ )} = 0

Simplify

csc (2 χ) - \frac{1}{1 - cos ( 2 χ )} : \frac{csc ( 2 χ ) ( 1 - cos ( 2 χ ) ) - 1}{1 - cos ( 2 χ )}

csc (2 χ) - \frac{1}{1 - cos ( 2 χ )}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

\frac{csc ( 2 χ ) ( 1 - cos ( 2 χ ) ) - 1}{1 - cos ( 2 χ )} = 0

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

csc (2 χ) (1 - cos (2 χ)) - 1 = 0

Express with sin, cos

\frac{1}{sin ( 2 χ )} (1 - cos (2 χ)) - 1 = 0

Simplify

\frac{1}{sin ( 2 χ )} (1 - cos (2 χ)) - 1 : \frac{1 - cos ( 2 χ ) - sin ( 2 χ )}{sin ( 2 χ )}

\frac{1}{sin ( 2 χ )} (1 - cos (2 χ)) - 1

\frac{1}{sin ( 2 χ )} (1 - cos (2 χ)) = \frac{1 - cos ( 2 χ )}{sin ( 2 χ )}

\frac{1}{sin ( 2 χ )} (1 - cos (2 χ))

Multiply fractions:

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

= \frac{1 \cdot ( 1 - cos ( 2 χ ) )}{sin ( 2 χ )}

1 \cdot (1 - cos (2 χ)) = 1 - cos (2 χ)

1 \cdot (1 - cos (2 χ))

Multiply:

1 \cdot (1 - cos (2 χ)) = (1 - cos (2 χ))

= (1 - cos (2 χ))

Remove parentheses:

(a) = a

= 1 - cos (2 χ)

= \frac{1 - cos ( 2 χ )}{sin ( 2 χ )}

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

Convert element to fraction:

1 = \frac{1 sin ( 2 χ )}{sin ( 2 χ )}

= \frac{1 - cos ( 2 χ )}{sin ( 2 χ )} - \frac{1 \cdot sin ( 2 χ )}{sin ( 2 χ )}

Since the denominators are equal, combine the fractions:

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

= \frac{1 - cos ( 2 χ ) - 1 \cdot sin ( 2 χ )}{sin ( 2 χ )}

Multiply:

1 \cdot sin (2 χ) = sin (2 χ)

= \frac{1 - cos ( 2 χ ) - sin ( 2 χ )}{sin ( 2 χ )}

\frac{1 - cos ( 2 χ ) - sin ( 2 χ )}{sin ( 2 χ )} = 0

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

1 - cos (2 χ) - sin (2 χ) = 0

Add

sin (2 χ)

to both sides

1 - cos (2 χ) = sin (2 χ)

Square both sides

(1 - cos (2 χ))^{2} = sin^{2} (2 χ)

Subtract

sin^{2} (2 χ)

from both sides

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

Factor

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

(1 - cos (2 χ))^{2} - sin^{2} (2 χ)

Apply Difference of Two Squares Formula:

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

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

= ((1 - cos (2 χ)) + sin (2 χ)) ((1 - cos (2 χ)) - sin (2 χ))

Refine

= (sin (2 χ) - cos (2 χ) + 1) (- cos (2 χ) - sin (2 χ) + 1)

(1 - cos (2 χ) + sin (2 χ)) (1 - cos (2 χ) - sin (2 χ)) = 0

Solving each part separately

1 - cos (2 χ) + sin (2 χ) = 0 or 1 - cos (2 χ) - sin (2 χ) = 0

1 - cos (2 χ) + sin (2 χ) = 0 : χ = πn + \frac{3 π}{4}, χ = πn + π

1 - cos (2 χ) + sin (2 χ) = 0

Rewrite using trig identities

1 - cos (2 χ) + sin (2 χ)

sin (2 χ) - cos (2 χ) = 2 sin (2 χ - \frac{π}{4})

sin (2 χ) - cos (2 χ)

Rewrite as

= 2 (\frac{1}{2} sin (2 χ) - \frac{1}{2} cos (2 χ))

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{1}{2}

Use the following trivial identity:

sin (\frac{π}{4}) = \frac{1}{2}

= 2 (cos (\frac{π}{4}) sin (2 χ) - sin (\frac{π}{4}) cos (2 χ))

Use the Angle Sum identity:

sin (s - t) = sin (s) cos (t) - cos (s) sin (t)

= 2 sin (2 χ - \frac{π}{4})

= 1 + 2 sin (2 χ - \frac{π}{4})

1 + 2 sin (2 χ - \frac{π}{4}) = 0

Move

1

to the right side

1 + 2 sin (2 χ - \frac{π}{4}) = 0

Subtract

1

from both sides

1 + 2 sin (2 χ - \frac{π}{4}) - 1 = 0 - 1

Simplify

2 sin (2 χ - \frac{π}{4}) = - 1

2 sin (2 χ - \frac{π}{4}) = - 1

Divide both sides by

2

2 sin (2 χ - \frac{π}{4}) = - 1

Divide both sides by

2

\frac{2 sin ( 2 χ - \frac{π}{4} )}{2} = \frac{- 1}{2}

Simplify

\frac{2 sin ( 2 χ - \frac{π}{4} )}{2} = \frac{- 1}{2}

Simplify

\frac{2 sin ( 2 χ - \frac{π}{4} )}{2} : sin (2 χ - \frac{π}{4})

\frac{2 sin ( 2 χ - \frac{π}{4} )}{2}

Cancel the common factor:

2

= sin (2 χ - \frac{π}{4})

Simplify

\frac{- 1}{2} : - \frac{2}{2}

\frac{- 1}{2}

Apply the fraction rule:

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

= - \frac{1}{2}

Rationalize

- \frac{1}{2} : - \frac{2}{2}

- \frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2}

sin (2 χ - \frac{π}{4}) = - \frac{2}{2}

sin (2 χ - \frac{π}{4}) = - \frac{2}{2}

sin (2 χ - \frac{π}{4}) = - \frac{2}{2}

General solutions for

sin (2 χ - \frac{π}{4}) = - \frac{2}{2}

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}

2 χ - \frac{π}{4} = \frac{5 π}{4} + 2 πn, 2 χ - \frac{π}{4} = \frac{7 π}{4} + 2 πn

2 χ - \frac{π}{4} = \frac{5 π}{4} + 2 πn, 2 χ - \frac{π}{4} = \frac{7 π}{4} + 2 πn

Solve

2 χ - \frac{π}{4} = \frac{5 π}{4} + 2 πn : χ = πn + \frac{3 π}{4}

2 χ - \frac{π}{4} = \frac{5 π}{4} + 2 πn

Move

\frac{π}{4}

to the right side

2 χ - \frac{π}{4} = \frac{5 π}{4} + 2 πn

Add

\frac{π}{4}

to both sides

2 χ - \frac{π}{4} + \frac{π}{4} = \frac{5 π}{4} + 2 πn + \frac{π}{4}

Simplify

2 χ - \frac{π}{4} + \frac{π}{4} = \frac{5 π}{4} + 2 πn + \frac{π}{4}

Simplify

2 χ - \frac{π}{4} + \frac{π}{4} : 2 χ

2 χ - \frac{π}{4} + \frac{π}{4}

Add similar elements:

- \frac{π}{4} + \frac{π}{4} = 0

= 2 χ

Simplify

\frac{5 π}{4} + 2 πn + \frac{π}{4} : 2 πn + \frac{3 π}{2}

\frac{5 π}{4} + 2 πn + \frac{π}{4}

Group like terms

= 2 πn + \frac{π}{4} + \frac{5 π}{4}

Combine the fractions

\frac{π}{4} + \frac{5 π}{4} : \frac{3 π}{2}

Apply rule

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

= \frac{π + 5 π}{4}

Add similar elements:

π + 5 π = 6 π

= \frac{6 π}{4}

Cancel the common factor:

2

= \frac{3 π}{2}

= 2 πn + \frac{3 π}{2}

2 χ = 2 πn + \frac{3 π}{2}

2 χ = 2 πn + \frac{3 π}{2}

2 χ = 2 πn + \frac{3 π}{2}

Divide both sides by

2

2 χ = 2 πn + \frac{3 π}{2}

Divide both sides by

2

\frac{2 χ}{2} = \frac{2 πn}{2} + \frac{\frac{3 π}{2}}{2}

Simplify

\frac{2 χ}{2} = \frac{2 πn}{2} + \frac{\frac{3 π}{2}}{2}

Simplify

\frac{2 χ}{2} : χ

\frac{2 χ}{2}

Divide the numbers:

\frac{2}{2} = 1

= χ

Simplify

\frac{2 πn}{2} + \frac{\frac{3 π}{2}}{2} : πn + \frac{3 π}{4}

\frac{2 πn}{2} + \frac{\frac{3 π}{2}}{2}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

\frac{\frac{3 π}{2}}{2} = \frac{3 π}{4}

\frac{\frac{3 π}{2}}{2}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 π}{4}

= πn + \frac{3 π}{4}

χ = πn + \frac{3 π}{4}

χ = πn + \frac{3 π}{4}

χ = πn + \frac{3 π}{4}

Solve

2 χ - \frac{π}{4} = \frac{7 π}{4} + 2 πn : χ = πn + π

2 χ - \frac{π}{4} = \frac{7 π}{4} + 2 πn

Move

\frac{π}{4}

to the right side

2 χ - \frac{π}{4} = \frac{7 π}{4} + 2 πn

Add

\frac{π}{4}

to both sides

2 χ - \frac{π}{4} + \frac{π}{4} = \frac{7 π}{4} + 2 πn + \frac{π}{4}

Simplify

2 χ - \frac{π}{4} + \frac{π}{4} = \frac{7 π}{4} + 2 πn + \frac{π}{4}

Simplify

2 χ - \frac{π}{4} + \frac{π}{4} : 2 χ

2 χ - \frac{π}{4} + \frac{π}{4}

Add similar elements:

- \frac{π}{4} + \frac{π}{4} = 0

= 2 χ

Simplify

\frac{7 π}{4} + 2 πn + \frac{π}{4} : 2 πn + 2 π

\frac{7 π}{4} + 2 πn + \frac{π}{4}

Group like terms

= 2 πn + \frac{π}{4} + \frac{7 π}{4}

Combine the fractions

\frac{π}{4} + \frac{7 π}{4} : 2 π

Apply rule

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

= \frac{π + 7 π}{4}

Add similar elements:

π + 7 π = 8 π

= \frac{8 π}{4}

Divide the numbers:

\frac{8}{4} = 2

= 2 π

= 2 πn + 2 π

2 χ = 2 πn + 2 π

2 χ = 2 πn + 2 π

2 χ = 2 πn + 2 π

Divide both sides by

2

2 χ = 2 πn + 2 π

Divide both sides by

2

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

Simplify

χ = πn + π

χ = πn + π

χ = πn + \frac{3 π}{4}, χ = πn + π

1 - cos (2 χ) - sin (2 χ) = 0 : χ = 2 πn, χ = π + 2 πn, χ = \frac{π}{4} + πn

1 - cos (2 χ) - sin (2 χ) = 0

Rewrite using trig identities

1 - cos (2 χ) - sin (2 χ)

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

= 1 - cos (2 χ) - 2 sin (χ) cos (χ)

Use the Double Angle identity:

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

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

Simplify

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

1 - (1 - 2 sin^{2} (χ)) - 2 cos (χ) sin (χ)

- (1 - 2 sin^{2} (χ)) : - 1 + 2 sin^{2} (χ)

- (1 - 2 sin^{2} (χ))

Distribute parentheses

= - (1) - (- 2 sin^{2} (χ))

Apply minus-plus rules

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

= - 1 + 2 sin^{2} (χ)

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

1 - 1 = 0

= 2 sin^{2} (χ) - 2 cos (χ) sin (χ)

= 2 sin^{2} (χ) - 2 cos (χ) sin (χ)

2 sin^{2} (χ) - 2 cos (χ) sin (χ) = 0

Factor

2 sin^{2} (χ) - 2 cos (χ) sin (χ) : 2 sin (χ) (sin (χ) - cos (χ))

2 sin^{2} (χ) - 2 cos (χ) sin (χ)

Apply exponent rule:

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

sin^{2} (χ) = sin (χ) sin (χ)

= 2 sin (χ) sin (χ) - 2 sin (χ) cos (χ)

Factor out common term

2 sin (χ)

= 2 sin (χ) (sin (χ) - cos (χ))

2 sin (χ) (sin (χ) - cos (χ)) = 0

Solving each part separately

sin (χ) = 0 or sin (χ) - cos (χ) = 0

sin (χ) = 0 : χ = 2 πn, χ = π + 2 πn

sin (χ) = 0

General solutions for

sin (χ) = 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}

χ = 0 + 2 πn, χ = π + 2 πn

χ = 0 + 2 πn, χ = π + 2 πn

Solve

χ = 0 + 2 πn : χ = 2 πn

χ = 0 + 2 πn

0 + 2 πn = 2 πn

χ = 2 πn

χ = 2 πn, χ = π + 2 πn

sin (χ) - cos (χ) = 0 : χ = \frac{π}{4} + πn

sin (χ) - cos (χ) = 0

Rewrite using trig identities

sin (χ) - cos (χ) = 0

Divide both sides by

cos (χ), cos (χ) \neq = 0

\frac{sin ( χ ) - cos ( χ )}{cos ( χ )} = \frac{0}{cos ( χ )}

Simplify

\frac{sin ( χ )}{cos ( χ )} - 1 = 0

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

tan (χ) - 1 = 0

tan (χ) - 1 = 0

Move

1

to the right side

tan (χ) - 1 = 0

Add

1

to both sides

tan (χ) - 1 + 1 = 0 + 1

Simplify

tan (χ) = 1

tan (χ) = 1

General solutions for

tan (χ) = 1

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

χ = \frac{π}{4} + πn

χ = \frac{π}{4} + πn

Combine all the solutions

χ = 2 πn, χ = π + 2 πn, χ = \frac{π}{4} + πn

Combine all the solutions

χ = πn + \frac{3 π}{4}, χ = πn + π, χ = 2 πn, χ = π + 2 πn, χ = \frac{π}{4} + πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

csc (2 χ) = \frac{1}{1 - cos ( 2 χ )}

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

Check the solution

πn + \frac{3 π}{4} :

False

πn + \frac{3 π}{4}

Plug in

n = 1

π 1 + \frac{3 π}{4}

For

csc (2 χ) = \frac{1}{1 - cos ( 2 χ )}

plug in

χ = π 1 + \frac{3 π}{4}

csc (2 (π 1 + \frac{3 π}{4})) = \frac{1}{1 - cos ( 2 ( π 1 + \frac{3 π}{4} ) )}

Refine

- 1 = 1

\Rightarrow False

Check the solution

πn + π :

False

πn + π

Plug in

n = 1

π 1 + π

For

csc (2 χ) = \frac{1}{1 - cos ( 2 χ )}

plug in

χ = π 1 + π

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

Undefined

\Rightarrow False

Check the solution

2 πn :

False

2 πn

Plug in

n = 1

2 π 1

For

csc (2 χ) = \frac{1}{1 - cos ( 2 χ )}

plug in

χ = 2 π 1

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

csc (2 χ) = \frac{1}{1 - cos ( 2 χ )}

plug in

χ = π + 2 π 1

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

Undefined

\Rightarrow False

Check the solution

\frac{π}{4} + πn :

True

\frac{π}{4} + πn

Plug in

n = 1

\frac{π}{4} + π 1

For

csc (2 χ) = \frac{1}{1 - cos ( 2 χ )}

plug in

χ = \frac{π}{4} + π 1

csc (2 (\frac{π}{4} + π 1)) = \frac{1}{1 - cos ( 2 ( \frac{π}{4} + π 1 ) )}

Refine

1 = 1

\Rightarrow True

χ = \frac{π}{4} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for csc(2χ)= 1/(1-cos(2χ)) ?
The general solution for csc(2χ)= 1/(1-cos(2χ)) is χ= pi/4+pin