What is the general solution for 5cot^2(x)-15=0 ?

The general solution for 5cot^2(x)-15=0 is x= pi/6+pin,x=(5pi)/6+pin

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5cot^2(x)-15=0

5 cot^{2} (x) - 15 = 0

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

5 cot^{2} (x) - 15 = 0

Solve by substitution

5 cot^{2} (x) - 15 = 0

Let:

cot (x) = u

5 u^{2} - 15 = 0

5 u^{2} - 15 = 0 : u = 3, u = - 3

5 u^{2} - 15 = 0

Move

15

to the right side

5 u^{2} - 15 = 0

Add

15

to both sides

5 u^{2} - 15 + 15 = 0 + 15

Simplify

5 u^{2} = 15

5 u^{2} = 15

Divide both sides by

5

5 u^{2} = 15

Divide both sides by

5

\frac{5 u ^{2}}{5} = \frac{15}{5}

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

sin (x) = \frac{2}{5}

csc (x) = 5

sec^{2} (x) + tan^{2} (x) - 1 = 0

tan^{2} (x) - 3 tan (x) = 0, 0 \leq x < 2 π

tan (θ) = 10

What is the general solution for 5cot^2(x)-15=0 ?
The general solution for 5cot^2(x)-15=0 is x= pi/6+pin,x=(5pi)/6+pin