What is the general solution for cot^2(x)-7cot(x)+10=0 ?

The general solution for cot^2(x)-7cot(x)+10=0 is x=0.19739…+pin,x=0.46364…+pin

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cot^2(x)-7cot(x)+10=0

Solution

cot^{2} (x) - 7 cot (x) + 10 = 0

Solution

x = 0.19739 \dots + πn, x = 0.46364 \dots + πn

Degrees

x = 11.30993 \dots^{\circ} + 18 0^{\circ} n, x = 26.56505 \dots^{\circ} + 18 0^{\circ} n

Solution steps

cot^{2} (x) - 7 cot (x) + 10 = 0

Solve by substitution

cot^{2} (x) - 7 cot (x) + 10 = 0

Let:

cot (x) = u

u^{2} - 7 u + 10 = 0

u^{2} - 7 u + 10 = 0 : u = 5, u = 2

u^{2} - 7 u + 10 = 0

Solve with the quadratic formula

u^{2} - 7 u + 10 = 0

Quadratic Equation Formula:

For

a = 1, b = - 7, c = 10

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot 10}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot 10}{2 \cdot 1}

(- 7)^{2} - 4 \cdot 1 \cdot 10 = 3

(- 7)^{2} - 4 \cdot 1 \cdot 10

Apply exponent rule:

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

n

is even

(- 7)^{2} = 7^{2}

= 7^{2} - 4 \cdot 1 \cdot 10

Multiply the numbers:

4 \cdot 1 \cdot 10 = 40

= 7^{2} - 40

7^{2} = 49

= 49 - 40

Subtract the numbers:

49 - 40 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- ( - 7 ) \pm 3}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 7 ) + 3}{2 \cdot 1}, u_{2} = \frac{- ( - 7 ) - 3}{2 \cdot 1}

u = \frac{- ( - 7 ) + 3}{2 \cdot 1} : 5

\frac{- ( - 7 ) + 3}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{7 + 3}{2 \cdot 1}

Add the numbers:

7 + 3 = 10

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{10}{2}

Divide the numbers:

\frac{10}{2} = 5

= 5

u = \frac{- ( - 7 ) - 3}{2 \cdot 1} : 2

\frac{- ( - 7 ) - 3}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{7 - 3}{2 \cdot 1}

Subtract the numbers:

7 - 3 = 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 = 5, u = 2

Substitute back

u = cot (x)

cot (x) = 5, cot (x) = 2

cot (x) = 5, cot (x) = 2

cot (x) = 5 : x = arccot (5) + πn

cot (x) = 5

Apply trig inverse properties

cot (x) = 5

General solutions for

cot (x) = 5

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

x = arccot (5) + πn

x = arccot (5) + πn

cot (x) = 2 : x = arccot (2) + πn

cot (x) = 2

Apply trig inverse properties

cot (x) = 2

General solutions for

cot (x) = 2

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

x = arccot (2) + πn

x = arccot (2) + πn

Combine all the solutions

x = arccot (5) + πn, x = arccot (2) + πn

Show solutions in decimal form

x = 0.19739 \dots + πn, x = 0.46364 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for cot^2(x)-7cot(x)+10=0 ?
The general solution for cot^2(x)-7cot(x)+10=0 is x=0.19739…+pin,x=0.46364…+pin