What is the general solution for cot^2(θ)-cot(θ)=0 ?

The general solution for cot^2(θ)-cot(θ)=0 is θ= pi/4+pin,θ= pi/2+pin

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cot^2(θ)-cot(θ)=0

Solution

cot^{2} (θ) - cot (θ) = 0

Solution

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

Degrees

θ = 4 5^{\circ} + 18 0^{\circ} n, θ = 9 0^{\circ} + 18 0^{\circ} n

Solution steps

cot^{2} (θ) - cot (θ) = 0

Solve by substitution

cot^{2} (θ) - cot (θ) = 0

Let:

cot (θ) = u

u^{2} - u = 0

u^{2} - u = 0 : u = 1, u = 0

u^{2} - u = 0

Solve with the quadratic formula

u^{2} - u = 0

Quadratic Equation Formula:

For

a = 1, b = - 1, c = 0

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot 0 = 1

(- 1)^{2} - 4 \cdot 1 \cdot 0

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 1 \cdot 0 = 0

4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 0

= 1 - 0

Subtract the numbers:

1 - 0 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- ( - 1 ) \pm 1}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 1}{2 \cdot 1}, u_{2} = \frac{- ( - 1 ) - 1}{2 \cdot 1}

u = \frac{- ( - 1 ) + 1}{2 \cdot 1} : 1

\frac{- ( - 1 ) + 1}{2 \cdot 1}

Apply rule

- (- a) = a

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

Add the numbers:

1 + 1 = 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 1 ) - 1}{2 \cdot 1} : 0

\frac{- ( - 1 ) - 1}{2 \cdot 1}

Apply rule

- (- a) = a

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

Subtract the numbers:

1 - 1 = 0

= \frac{0}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

The solutions to the quadratic equation are:

u = 1, u = 0

Substitute back

u = cot (θ)

cot (θ) = 1, cot (θ) = 0

cot (θ) = 1, cot (θ) = 0

cot (θ) = 1 : θ = \frac{π}{4} + πn

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{π}{4} + πn

θ = \frac{π}{4} + πn

cot (θ) = 0 : θ = \frac{π}{2} + πn

cot (θ) = 0

General solutions for

cot (θ) = 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

θ = \frac{π}{2} + πn

θ = \frac{π}{2} + πn

Combine all the solutions

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

Popular Examples

5cos(2θ)=-15cos(θ)-10

5 cos (2 θ) = - 15 cos (θ) - 10

2cos(x)-5=3sec(x)

2 cos (x) - 5 = 3 sec (x)

sin(3θ+pi/(18))=1,0<= θ<2pi

sin (3 θ + \frac{π}{18}) = 1, 0 \leq θ < 2 π

1=tan(c)

1 = tan (c)

sin(a)=(sqrt(3))/2

sin (a) = \frac{3}{2}

Frequently Asked Questions (FAQ)

What is the general solution for cot^2(θ)-cot(θ)=0 ?
The general solution for cot^2(θ)-cot(θ)=0 is θ= pi/4+pin,θ= pi/2+pin