What is the general solution for cot(x)cos^2(x)=cot(x) ?

The general solution for cot(x)cos^2(x)=cot(x) is x= pi/2+pin

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cot(x)cos^2(x)=cot(x)

Solution

cot (x) cos^{2} (x) = cot (x)

Solution

x = \frac{π}{2} + πn

Degrees

x = 9 0^{\circ} + 18 0^{\circ} n

Solution steps

cot (x) cos^{2} (x) = cot (x)

Subtract

cot (x)

from both sides

cos^{2} (x) cot (x) - cot (x) = 0

Factor

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

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

Factor out common term

cot (x)

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

Factor

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

cos^{2} (x) - 1

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

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

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

= (cos (x) + 1) (cos (x) - 1)

= cot (x) (cos (x) + 1) (cos (x) - 1)

cot (x) (cos (x) + 1) (cos (x) - 1) = 0

Solving each part separately

cot (x) = 0 or cos (x) + 1 = 0 or cos (x) - 1 = 0

cot (x) = 0 : x = \frac{π}{2} + πn

cot (x) = 0

General solutions for

cot (x) = 0

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

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

cos (x) + 1 = 0 : x = π + 2 πn

cos (x) + 1 = 0

Move

1

to the right side

cos (x) + 1 = 0

Subtract

1

from both sides

cos (x) + 1 - 1 = 0 - 1

Simplify

cos (x) = - 1

cos (x) = - 1

General solutions for

cos (x) = - 1

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}

x = π + 2 πn

x = π + 2 πn

cos (x) - 1 = 0 : x = 2 πn

cos (x) - 1 = 0

Move

1

to the right side

cos (x) - 1 = 0

Add

1

to both sides

cos (x) - 1 + 1 = 0 + 1

Simplify

cos (x) = 1

cos (x) = 1

General solutions for

cos (x) = 1

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}

x = 0 + 2 πn

x = 0 + 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

Combine all the solutions

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

Since the equation is undefined for:

π + 2 πn, 2 πn

x = \frac{π}{2} + πn

Popular Examples

2sin(2x)+5=4

2 sin (2 x) + 5 = 4

solvefor x,sin(x)=cos(x)solve for

x, sin (x) = cos (x)

sin(θ)=-(sqrt(3))/2 ,0<= θ<= 2pi

sin (θ) = - \frac{3}{2}, 0 \leq θ \leq 2 π

sin(x)+cos(x)= 1/(sqrt(2))

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

cot(x)=4

cot (x) = 4

Frequently Asked Questions (FAQ)

What is the general solution for cot(x)cos^2(x)=cot(x) ?
The general solution for cot(x)cos^2(x)=cot(x) is x= pi/2+pin