What is the general solution for cos(θ)=cot(θ) ?

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

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Popular >

cos(θ)=cot(θ)

Solution

cos (θ) = cot (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

cos (θ) = cot (θ)

Subtract

cot (θ)

from both sides

cos (θ) - cot (θ) = 0

Express with sin, cos

cos (θ) - cot (θ)

Use the basic trigonometric identity:

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

= cos (θ) - \frac{cos ( θ )}{sin ( θ )}

Simplify

cos (θ) - \frac{cos ( θ )}{sin ( θ )} : \frac{cos ( θ ) sin ( θ ) - cos ( θ )}{sin ( θ )}

cos (θ) - \frac{cos ( θ )}{sin ( θ )}

Convert element to fraction:

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

= \frac{cos ( θ ) sin ( θ )}{sin ( θ )} - \frac{cos ( θ )}{sin ( θ )}

Since the denominators are equal, combine the fractions:

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

= \frac{cos ( θ ) sin ( θ ) - cos ( θ )}{sin ( θ )}

= \frac{cos ( θ ) sin ( θ ) - cos ( θ )}{sin ( θ )}

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

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

- cos (θ) + cos (θ) sin (θ) = 0

Factor

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

- cos (θ) + cos (θ) sin (θ)

Factor out common term

cos (θ)

= cos (θ) (- 1 + sin (θ))

cos (θ) (sin (θ) - 1) = 0

Solving each part separately

cos (θ) = 0 or sin (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

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

sin (θ) - 1 = 0

Move

1

to the right side

sin (θ) - 1 = 0

Add

1

to both sides

sin (θ) - 1 + 1 = 0 + 1

Simplify

sin (θ) = 1

sin (θ) = 1

General solutions for

sin (θ) = 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}

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

Popular Examples

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sin(x)=-4/5 ,cos(x)<0,cos(x/2)

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1+sin(x)-2sin^2(x)=0

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

solvefor y,z=e^{-x}sin(x+2y)solve for

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3 sin (θ) - 2 sin^{2} (θ) = 0

Frequently Asked Questions (FAQ)

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