What is the general solution for 19cot^2(x)-19=0 ?

The general solution for 19cot^2(x)-19=0 is x= pi/4+pin,x=(3pi)/4+pin

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19cot^2(x)-19=0

Solution

19 cot^{2} (x) - 19 = 0

Solution

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

19 cot^{2} (x) - 19 = 0

Solve by substitution

19 cot^{2} (x) - 19 = 0

Let:

cot (x) = u

19 u^{2} - 19 = 0

19 u^{2} - 19 = 0 : u = 1, u = - 1

19 u^{2} - 19 = 0

Move

19

to the right side

19 u^{2} - 19 = 0

Add

19

to both sides

19 u^{2} - 19 + 19 = 0 + 19

Simplify

19 u^{2} = 19

19 u^{2} = 19

Divide both sides by

19

19 u^{2} = 19

Divide both sides by

19

\frac{19 u ^{2}}{19} = \frac{19}{19}

Simplify

u^{2} = 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

Substitute back

u = cot (x)

cot (x) = 1, cot (x) = - 1

cot (x) = 1, cot (x) = - 1

cot (x) = 1 : x = \frac{π}{4} + πn

cot (x) = 1

General solutions for

cot (x) = 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

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

cot (x) = - 1 : x = \frac{3 π}{4} + πn

cot (x) = - 1

General solutions for

cot (x) = - 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

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

Combine all the solutions

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 19cot^2(x)-19=0 ?
The general solution for 19cot^2(x)-19=0 is x= pi/4+pin,x=(3pi)/4+pin