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

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

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cot^2(a)=1

cot^{2} (a) = 1

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

Degrees

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

Solution steps

cot^{2} (a) = 1

Solve by substitution

cot^{2} (a) = 1

Let:

cot (a) = 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 (a)

cot (a) = 1, cot (a) = - 1

cot (a) = 1, cot (a) = - 1

cot (a) = 1 : a = \frac{π}{4} + πn

cot (a) = 1

General solutions for

cot (a) = 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

a = \frac{π}{4} + πn

a = \frac{π}{4} + πn

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

cot (a) = - 1

General solutions for

cot (a) = - 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

a = \frac{3 π}{4} + πn

a = \frac{3 π}{4} + πn

Combine all the solutions

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

3 sec^{2} (x) - 4 = 0, 0 \leq x < 2 π

2 sec (θ) + 4 = 0

sin (x) = \frac{( 60 \cdot sin ( 15 5 ^{\circ} ) )}{97.72}

(1 - 2 cos (x)) (1 + sin (x)) = 0

3 cos (2 x) + 7 sin (x) - 5 = 0

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