What is the general solution for 3tan^2(x)-9=0 ?

The general solution for 3tan^2(x)-9=0 is x= pi/3+pin,x=(2pi)/3+pin

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3tan^2(x)-9=0

3 tan^{2} (x) - 9 = 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

3 tan^{2} (x) - 9 = 0

Solve by substitution

3 tan^{2} (x) - 9 = 0

Let:

tan (x) = u

3 u^{2} - 9 = 0

3 u^{2} - 9 = 0 : u = 3, u = - 3

3 u^{2} - 9 = 0

Move

9

to the right side

3 u^{2} - 9 = 0

Add

9

to both sides

3 u^{2} - 9 + 9 = 0 + 9

Simplify

3 u^{2} = 9

3 u^{2} = 9

Divide both sides by

3

3 u^{2} = 9

Divide both sides by

3

\frac{3 u ^{2}}{3} = \frac{9}{3}

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

3 cos (4 x) = 2

2 cos^{2} (x) = cos (x) + 1

4 cos (θ) + 2 = 0

6 cos^{2} (θ) = 3

cos (x) + cos (3 x) = 0

What is the general solution for 3tan^2(x)-9=0 ?
The general solution for 3tan^2(x)-9=0 is x= pi/3+pin,x=(2pi)/3+pin