What is the general solution for cot^2(θ)=cot(θ),0<= θ<= 2pi ?

Q: What is the general solution for cot^2(θ)=cot(θ),0<= θ<= 2pi ?

The general solution for cot^2(θ)=cot(θ),0<= θ<= 2pi is θ= pi/4 ,θ=(5pi)/4 ,θ= pi/2 ,θ=(3pi)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cot^2(θ)=cot(θ),0<= θ<= 2pi

Solution

cot^{2} (θ) = cot (θ), 0 \leq θ \leq 2 π

Solution

θ = \frac{π}{4}, θ = \frac{5 π}{4}, θ = \frac{π}{2}, θ = \frac{3 π}{2}

Degrees

θ = 4 5^{\circ}, θ = 22 5^{\circ}, θ = 9 0^{\circ}, θ = 27 0^{\circ}

Solution steps

cot^{2} (θ) = cot (θ), 0 \leq θ \leq 2 π

Solve by substitution

cot^{2} (θ) = cot (θ)

Let:

cot (θ) = u

u^{2} = u

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

u^{2} = u

Move

u

to the left side

u^{2} = u

Subtract

u

from both sides

u^{2} - u = u - u

Simplify

u^{2} - 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, 0 \leq θ \leq 2 π : θ = \frac{π}{4}, θ = \frac{5 π}{4}

cot (θ) = 1, 0 \leq θ \leq 2 π

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

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{π}{4}, θ = \frac{5 π}{4}

cot (θ) = 0, 0 \leq θ \leq 2 π : θ = \frac{π}{2}, θ = \frac{3 π}{2}

cot (θ) = 0, 0 \leq θ \leq 2 π

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

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{π}{2}, θ = \frac{3 π}{2}

Combine all the solutions

θ = \frac{π}{4}, θ = \frac{5 π}{4}, θ = \frac{π}{2}, θ = \frac{3 π}{2}

Popular Examples

cos(x)= 9/19

cos (x) = \frac{9}{19}

cos(α)=(sqrt(5))/3

cos (α) = \frac{5}{3}

2-3sin(x)-cos(2x)=0

2 - 3 sin (x) - cos (2 x) = 0

cos(2θ)= 1/4

cos (2 θ) = \frac{1}{4}

sec(θ)= 25/7

sec (θ) = \frac{25}{7}

Frequently Asked Questions (FAQ)

What is the general solution for cot^2(θ)=cot(θ),0<= θ<= 2pi ?
The general solution for cot^2(θ)=cot(θ),0<= θ<= 2pi is θ= pi/4 ,θ=(5pi)/4 ,θ= pi/2 ,θ=(3pi)/2