What is the general solution for cot^2(θ)+cot(θ)-20=0 ?

The general solution for cot^2(θ)+cot(θ)-20=0 is θ=0.24497…+pin,θ=2.94419…+pin

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Popular >

cot^2(θ)+cot(θ)-20=0

Solution

cot^{2} (θ) + cot (θ) - 20 = 0

Solution

θ = 0.24497 \dots + πn, θ = 2.94419 \dots + πn

Degrees

θ = 14.03624 \dots^{\circ} + 18 0^{\circ} n, θ = 168.69006 \dots^{\circ} + 18 0^{\circ} n

Solution steps

cot^{2} (θ) + cot (θ) - 20 = 0

Solve by substitution

cot^{2} (θ) + cot (θ) - 20 = 0

Let:

cot (θ) = u

u^{2} + u - 20 = 0

u^{2} + u - 20 = 0 : u = 4, u = - 5

u^{2} + u - 20 = 0

Solve with the quadratic formula

u^{2} + u - 20 = 0

Quadratic Equation Formula:

For

a = 1, b = 1, c = - 20

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 20 )}{2 \cdot 1}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 20 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 20) = 9

1^{2} - 4 \cdot 1 \cdot (- 20)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 20)

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 20

Multiply the numbers:

4 \cdot 1 \cdot 20 = 80

= 1 + 80

Add the numbers:

1 + 80 = 81

= 81

Factor the number:

81 = 9^{2}

= 9^{2}

Apply radical rule:

9^{2} = 9

= 9

u_{1, 2} = \frac{- 1 \pm 9}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 1 + 9}{2 \cdot 1}, u_{2} = \frac{- 1 - 9}{2 \cdot 1}

u = \frac{- 1 + 9}{2 \cdot 1} : 4

\frac{- 1 + 9}{2 \cdot 1}

Add/Subtract the numbers:

- 1 + 9 = 8

= \frac{8}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{8}{2}

Divide the numbers:

\frac{8}{2} = 4

= 4

u = \frac{- 1 - 9}{2 \cdot 1} : - 5

\frac{- 1 - 9}{2 \cdot 1}

Subtract the numbers:

- 1 - 9 = - 10

= \frac{- 10}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 10}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{10}{2}

Divide the numbers:

\frac{10}{2} = 5

= - 5

The solutions to the quadratic equation are:

u = 4, u = - 5

Substitute back

u = cot (θ)

cot (θ) = 4, cot (θ) = - 5

cot (θ) = 4, cot (θ) = - 5

cot (θ) = 4 : θ = arccot (4) + πn

cot (θ) = 4

Apply trig inverse properties

cot (θ) = 4

General solutions for

cot (θ) = 4

cot (x) = a \Rightarrow x = arccot (a) + πn

θ = arccot (4) + πn

θ = arccot (4) + πn

cot (θ) = - 5 : θ = arccot (- 5) + πn

cot (θ) = - 5

Apply trig inverse properties

cot (θ) = - 5

General solutions for

cot (θ) = - 5

cot (x) = - a \Rightarrow x = arccot (- a) + πn

θ = arccot (- 5) + πn

θ = arccot (- 5) + πn

Combine all the solutions

θ = arccot (4) + πn, θ = arccot (- 5) + πn

Show solutions in decimal form

θ = 0.24497 \dots + πn, θ = 2.94419 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for cot^2(θ)+cot(θ)-20=0 ?
The general solution for cot^2(θ)+cot(θ)-20=0 is θ=0.24497…+pin,θ=2.94419…+pin