What is the general solution for 2tan^2(x)-6=0 ?

The general solution for 2tan^2(x)-6=0 is x= pi/3+pin,x=(2pi)/3+pin

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2tan^2(x)-6=0

2 tan^{2} (x) - 6 = 0

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

2 tan^{2} (x) - 6 = 0

Solve by substitution

2 tan^{2} (x) - 6 = 0

Let:

tan (x) = u

2 u^{2} - 6 = 0

2 u^{2} - 6 = 0 : u = 3, u = - 3

2 u^{2} - 6 = 0

Move

6

to the right side

2 u^{2} - 6 = 0

Add

6

to both sides

2 u^{2} - 6 + 6 = 0 + 6

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 = 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

0 = 2 sin (x) - 1

t, x = cos (2 t)

- 9 sin^{2} (θ) - 3 cos (θ) + 2 = - 5

csc (x) = - 1 + cot (x)

2 sin (3 x) = 3

What is the general solution for 2tan^2(x)-6=0 ?
The general solution for 2tan^2(x)-6=0 is x= pi/3+pin,x=(2pi)/3+pin