What is the general solution for tan^2(x)-1=2 ?

The general solution for tan^2(x)-1=2 is x= pi/3+pin,x=(2pi)/3+pin

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tan^2(x)-1=2

tan^{2} (x) - 1 = 2

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

Degrees

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

Solution steps

tan^{2} (x) - 1 = 2

Solve by substitution

tan^{2} (x) - 1 = 2

Let:

tan (x) = u

u^{2} - 1 = 2

u^{2} - 1 = 2 : u = 3, u = - 3

u^{2} - 1 = 2

Move

1

to the right side

u^{2} - 1 = 2

Add

1

to both sides

u^{2} - 1 + 1 = 2 + 1

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 = tan (x)

tan (x) = 3, tan (x) = - 3

tan (x) = 3, tan (x) = - 3

tan (x) = 3 : x = \frac{π}{3} + πn

tan (x) = 3

General solutions for

tan (x) = 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}

x = \frac{π}{3} + πn

x = \frac{π}{3} + πn

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

tan (x) = - 3

General solutions for

tan (x) = - 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}

x = \frac{2 π}{3} + πn

x = \frac{2 π}{3} + πn

Combine all the solutions

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

cos (2 a) = \frac{2}{3}, sin (a)

cot (2 x - 3) - 1 = 0

4 sin^{32} (x) - 3 cos^{2} (x) = 0

4 tan^{2} (x) + 15 sec (x) = 0

sin^{4} (x) + cos^{4} (x) = cos^{4} (x)

What is the general solution for tan^2(x)-1=2 ?
The general solution for tan^2(x)-1=2 is x= pi/3+pin,x=(2pi)/3+pin