What is the general solution for 2^{cot^2(x)}=8 ?

The general solution for 2^{cot^2(x)}=8 is x= pi/6+pin,x=(5pi)/6+pin

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2^{cot^2(x)}=8

2^{c o t^{2} (x)} = 8

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^{c o t^{2} (x)} = 8

Solve by substitution

2^{c o t^{2} (x)} = 8

Let:

cot (x) = u

2^{u^{2}} = 8

2^{u^{2}} = 8 : u = 3, u = - 3

2^{u^{2}} = 8

Apply exponent rules

2^{u^{2}} = 8

Convert to base

2 : 2^{u^{2}} = 2^{3}

Convert

8

to base

2

8 = 2^{3}

2^{u^{2}} = 2^{3}

2^{u^{2}} = 2^{3}

a^{f (x)} = a^{g (x)}

, then

f (x) = g (x)

u^{2} = 3

u^{2} = 3

Solve

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

u^{2} = 3

For

x^{2} = f (a)

the solutions are

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

u = 3, u = - 3

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

2 \cdot cos^{2} (x) + cos (x) - 1 = 0

cos (π y) = 0

cos (t) = - \frac{1}{3}

3 cot (x) - 1 = 0, 0 \leq x \leq 2 π x

1 = cosh (x)

What is the general solution for 2^{cot^2(x)}=8 ?
The general solution for 2^{cot^2(x)}=8 is x= pi/6+pin,x=(5pi)/6+pin