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

The general solution for cos(θ)cot(θ)=-cos(θ) is θ= pi/2+2pin,θ=(3pi)/2+2pin,θ=(3pi)/4+pin

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cos(θ)cot(θ)=-cos(θ)

cos (θ) cot (θ) = - cos (θ)

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

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n, θ = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

cos (θ) cot (θ) = - cos (θ)

Subtract

- cos (θ)

from both sides

cos (θ) cot (θ) + cos (θ) = 0

Factor

cos (θ) cot (θ) + cos (θ) : cos (θ) (cot (θ) + 1)

cos (θ) cot (θ) + cos (θ)

Factor out common term

cos (θ)

= cos (θ) (cot (θ) + 1)

cos (θ) (cot (θ) + 1) = 0

Solving each part separately

cos (θ) = 0 or cot (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

cot (θ) + 1 = 0 : θ = \frac{3 π}{4} + πn

cot (θ) + 1 = 0

Move

1

to the right side

cot (θ) + 1 = 0

Subtract

1

from both sides

cot (θ) + 1 - 1 = 0 - 1

Simplify

cot (θ) = - 1

cot (θ) = - 1

General solutions for

cot (θ) = - 1

cot (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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

Combine all the solutions

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

cos (θ) = \frac{( - \frac{4}{5} )}{7 13}

tan (\frac{x}{2}) = - \frac{1}{3}

C, y = arcsin (C) e^{2}

sin (θ) = \frac{2 3}{3}

cos (x + 3 π) - cos (x) + 1 = 0

What is the general solution for cos(θ)cot(θ)=-cos(θ) ?
The general solution for cos(θ)cot(θ)=-cos(θ) is θ= pi/2+2pin,θ=(3pi)/2+2pin,θ=(3pi)/4+pin