What is the general solution for 2cot^2(x)-4=2 ?

The general solution for 2cot^2(x)-4=2 is x= pi/6+pin,x=(5pi)/6+pin

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2cot^2(x)-4=2

2 cot^{2} (x) - 4 = 2

x = \frac{π}{6} + πn, x = \frac{5 π}{6} + πn

Degrees

x = 3 0^{\circ} + 18 0^{\circ} n, x = 15 0^{\circ} + 18 0^{\circ} n

Solution steps

2 cot^{2} (x) - 4 = 2

Solve by substitution

2 cot^{2} (x) - 4 = 2

Let:

cot (x) = u

2 u^{2} - 4 = 2

2 u^{2} - 4 = 2 : u = 3, u = - 3

2 u^{2} - 4 = 2

Move

4

to the right side

2 u^{2} - 4 = 2

Add

4

to both sides

2 u^{2} - 4 + 4 = 2 + 4

Simplify

2 u^{2} = 6

2 u^{2} = 6

Divide both sides by

2

2 u^{2} = 6

Divide both sides by

2

\frac{2 u ^{2}}{2} = \frac{6}{2}

Simplify

u^{2} = 3

u^{2} = 3

For

x^{2} = f (a)

the solutions are

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

u = 3, u = - 3

Substitute back

u = cot (x)

cot (x) = 3, cot (x) = - 3

cot (x) = 3, cot (x) = - 3

cot (x) = 3 : x = \frac{π}{6} + πn

cot (x) = 3

General solutions for

cot (x) = 3

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}{6} + πn

x = \frac{π}{6} + πn

cot (x) = - 3 : x = \frac{5 π}{6} + πn

cot (x) = - 3

General solutions for

cot (x) = - 3

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{5 π}{6} + πn

x = \frac{5 π}{6} + πn

Combine all the solutions

x = \frac{π}{6} + πn, x = \frac{5 π}{6} + πn

4 sin^{2} (θ) - 3 = 0, 0 \leq θ \leq 2 π

4 sin (B) + 8 = 0

4 cos (x) - 5 = - 7

0 = 1 + 2 sin (\frac{x ^{2}}{2})

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

What is the general solution for 2cot^2(x)-4=2 ?
The general solution for 2cot^2(x)-4=2 is x= pi/6+pin,x=(5pi)/6+pin