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

The general solution for (1+cos(θ))/(csc(θ))=sin(θ)+cos(θ) is θ= pi/2+2pin,θ=(3pi)/2+2pin

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(1+cos(θ))/(csc(θ))=sin(θ)+cos(θ)

Solution

\frac{1 + cos ( θ )}{csc ( θ )} = sin (θ) + cos (θ)

Solution

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

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

\frac{1 + cos ( θ )}{csc ( θ )} = sin (θ) + cos (θ)

Subtract

sin (θ) + cos (θ)

from both sides

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

Simplify

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

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

Convert element to fraction:

sin (θ) = \frac{sin ( θ ) csc ( θ )}{csc ( θ )}, cos (θ) = \frac{cos ( θ ) csc ( θ )}{csc ( θ )}

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

Since the denominators are equal, combine the fractions:

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

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

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

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

1 + cos (θ) - sin (θ) csc (θ) - cos (θ) csc (θ) = 0

Rewrite using trig identities

1 + cos (θ) - cos (θ) csc (θ) - csc (θ) sin (θ)

Use the basic trigonometric identity:

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

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

Simplify

1 + cos (θ) - cos (θ) csc (θ) - csc (θ) \frac{1}{csc ( θ )} : cos (θ) - cos (θ) csc (θ)

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

csc (θ) \frac{1}{csc ( θ )} = 1

csc (θ) \frac{1}{csc ( θ )}

Multiply fractions:

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

= \frac{1 \cdot csc ( θ )}{csc ( θ )}

Cancel the common factor:

csc (θ)

= 1

= 1 + cos (θ) - cos (θ) csc (θ) - 1

Group like terms

= cos (θ) - cos (θ) csc (θ) + 1 - 1

1 - 1 = 0

= cos (θ) - cos (θ) csc (θ)

= cos (θ) - cos (θ) csc (θ)

cos (θ) - cos (θ) csc (θ) = 0

Factor

cos (θ) - cos (θ) csc (θ) : - cos (θ) (csc (θ) - 1)

cos (θ) - cos (θ) csc (θ)

Factor out common term

- cos (θ)

= - cos (θ) (- 1 + csc (θ))

- cos (θ) (csc (θ) - 1) = 0

Solving each part separately

cos (θ) = 0 or csc (θ) - 1 = 0

cos (θ) = 0 : θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

cos (θ) = 0

General solutions for

cos (θ) = 0

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

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

csc (θ) - 1 = 0 : θ = \frac{π}{2} + 2 πn

csc (θ) - 1 = 0

Move

1

to the right side

csc (θ) - 1 = 0

Add

1

to both sides

csc (θ) - 1 + 1 = 0 + 1

Simplify

csc (θ) = 1

csc (θ) = 1

General solutions for

csc (θ) = 1

csc (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} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 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} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 2

θ = \frac{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

Combine all the solutions

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

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for (1+cos(θ))/(csc(θ))=sin(θ)+cos(θ) ?
The general solution for (1+cos(θ))/(csc(θ))=sin(θ)+cos(θ) is θ= pi/2+2pin,θ=(3pi)/2+2pin