What is the general solution for tan^2(b)=3 ?

The general solution for tan^2(b)=3 is b= pi/3+pin,b=(2pi)/3+pin

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tan^2(b)=3

tan^{2} (b) = 3

b = \frac{π}{3} + πn, b = \frac{2 π}{3} + πn

Degrees

b = 6 0^{\circ} + 18 0^{\circ} n, b = 12 0^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (b) = 3

Solve by substitution

tan^{2} (b) = 3

Let:

tan (b) = u

u^{2} = 3

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

Substitute back

u = tan (b)

tan (b) = 3, tan (b) = - 3

tan (b) = 3, tan (b) = - 3

tan (b) = 3 : b = \frac{π}{3} + πn

tan (b) = 3

General solutions for

tan (b) = 3

tan (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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

b = \frac{π}{3} + πn

b = \frac{π}{3} + πn

tan (b) = - 3 : b = \frac{2 π}{3} + πn

tan (b) = - 3

General solutions for

tan (b) = - 3

tan (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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

b = \frac{2 π}{3} + πn

b = \frac{2 π}{3} + πn

Combine all the solutions

b = \frac{π}{3} + πn, b = \frac{2 π}{3} + πn

a, sin (a + \frac{b}{2}) = cos (\frac{c}{2})

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

cot (\frac{- x + 3 n}{4}) = 0

sin (x) + sin^{2} (x) + sin^{3} (x) = 1

\frac{cos ( a )}{1 + sin ( a )} = sec (a)

What is the general solution for tan^2(b)=3 ?
The general solution for tan^2(b)=3 is b= pi/3+pin,b=(2pi)/3+pin