What is the general solution for cot^2(x)=1 ?

The general solution for cot^2(x)=1 is x= pi/4+pin,x=(3pi)/4+pin

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cot^2(x)=1

cot^{2} (x) = 1

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

cot^{2} (x) = 1

Solve by substitution

cot^{2} (x) = 1

Let:

cot (x) = u

u^{2} = 1

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

2 tan (x) - sec^{2} (x) = 0

4 sin^{2} (2 x) + 2 = 3

- \frac{1}{4} sin (\frac{x}{2}) = 0

8 sin (π x) - \frac{1}{4} + 1 = 7

8 sin^{2} (u) + 6 sin (u) - 9 = 0

What is the general solution for cot^2(x)=1 ?
The general solution for cot^2(x)=1 is x= pi/4+pin,x=(3pi)/4+pin