What is the general solution for cot^2(θ)-6cot(θ)+8=0 ?

The general solution for cot^2(θ)-6cot(θ)+8=0 is θ=0.24497…+pin,θ=0.46364…+pin

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cot^2(θ)-6cot(θ)+8=0

Solution

cot^{2} (θ) - 6 cot (θ) + 8 = 0

Solution

θ = 0.24497 \dots + πn, θ = 0.46364 \dots + πn

Degrees

θ = 14.03624 \dots^{\circ} + 18 0^{\circ} n, θ = 26.56505 \dots^{\circ} + 18 0^{\circ} n

Solution steps

cot^{2} (θ) - 6 cot (θ) + 8 = 0

Solve by substitution

cot^{2} (θ) - 6 cot (θ) + 8 = 0

Let:

cot (θ) = u

u^{2} - 6 u + 8 = 0

u^{2} - 6 u + 8 = 0 : u = 4, u = 2

u^{2} - 6 u + 8 = 0

Solve with the quadratic formula

u^{2} - 6 u + 8 = 0

Quadratic Equation Formula:

For

a = 1, b = - 6, c = 8

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 8}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 8}{2 \cdot 1}

(- 6)^{2} - 4 \cdot 1 \cdot 8 = 2

(- 6)^{2} - 4 \cdot 1 \cdot 8

Apply exponent rule:

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

n

is even

(- 6)^{2} = 6^{2}

= 6^{2} - 4 \cdot 1 \cdot 8

Multiply the numbers:

4 \cdot 1 \cdot 8 = 32

= 6^{2} - 32

6^{2} = 36

= 36 - 32

Subtract the numbers:

36 - 32 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

u_{1, 2} = \frac{- ( - 6 ) \pm 2}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 6 ) + 2}{2 \cdot 1}, u_{2} = \frac{- ( - 6 ) - 2}{2 \cdot 1}

u = \frac{- ( - 6 ) + 2}{2 \cdot 1} : 4

\frac{- ( - 6 ) + 2}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{6 + 2}{2 \cdot 1}

Add the numbers:

6 + 2 = 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{- ( - 6 ) - 2}{2 \cdot 1} : 2

\frac{- ( - 6 ) - 2}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{6 - 2}{2 \cdot 1}

Subtract the numbers:

6 - 2 = 4

= \frac{4}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2

The solutions to the quadratic equation are:

u = 4, u = 2

Substitute back

u = cot (θ)

cot (θ) = 4, cot (θ) = 2

cot (θ) = 4, cot (θ) = 2

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) = 2 : θ = arccot (2) + πn

cot (θ) = 2

Apply trig inverse properties

cot (θ) = 2

General solutions for

cot (θ) = 2

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

θ = arccot (2) + πn

θ = arccot (2) + πn

Combine all the solutions

θ = arccot (4) + πn, θ = arccot (2) + πn

Show solutions in decimal form

θ = 0.24497 \dots + πn, θ = 0.46364 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for cot^2(θ)-6cot(θ)+8=0 ?
The general solution for cot^2(θ)-6cot(θ)+8=0 is θ=0.24497…+pin,θ=0.46364…+pin