What is the general solution for 2tan^2(3x)=6 ?

The general solution for 2tan^2(3x)=6 is x= pi/9+(pin)/3 ,x=(2pi)/9+(pin)/3

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2tan^2(3x)=6

Solution

2 tan^{2} (3 x) = 6

Solution

x = \frac{π}{9} + \frac{πn}{3}, x = \frac{2 π}{9} + \frac{πn}{3}

Degrees

x = 2 0^{\circ} + 6 0^{\circ} n, x = 4 0^{\circ} + 6 0^{\circ} n

Solution steps

2 tan^{2} (3 x) = 6

Solve by substitution

2 tan^{2} (3 x) = 6

Let:

tan (3 x) = u

2 u^{2} = 6

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

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 (3 x)

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

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

tan (3 x) = 3 : x = \frac{π}{9} + \frac{πn}{3}

tan (3 x) = 3

General solutions for

tan (3 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}

3 x = \frac{π}{3} + πn

3 x = \frac{π}{3} + πn

Solve

3 x = \frac{π}{3} + πn : x = \frac{π}{9} + \frac{πn}{3}

3 x = \frac{π}{3} + πn

Divide both sides by

3

3 x = \frac{π}{3} + πn

Divide both sides by

3

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

Simplify

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

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

\frac{\frac{π}{3}}{3} + \frac{πn}{3} : \frac{π}{9} + \frac{πn}{3}

\frac{\frac{π}{3}}{3} + \frac{πn}{3}

\frac{\frac{π}{3}}{3} = \frac{π}{9}

\frac{\frac{π}{3}}{3}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{π}{3 \cdot 3}

Multiply the numbers:

3 \cdot 3 = 9

= \frac{π}{9}

= \frac{π}{9} + \frac{πn}{3}

x = \frac{π}{9} + \frac{πn}{3}

x = \frac{π}{9} + \frac{πn}{3}

x = \frac{π}{9} + \frac{πn}{3}

x = \frac{π}{9} + \frac{πn}{3}

tan (3 x) = - 3 : x = \frac{2 π}{9} + \frac{πn}{3}

tan (3 x) = - 3

General solutions for

tan (3 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}

3 x = \frac{2 π}{3} + πn

3 x = \frac{2 π}{3} + πn

Solve

3 x = \frac{2 π}{3} + πn : x = \frac{2 π}{9} + \frac{πn}{3}

3 x = \frac{2 π}{3} + πn

Divide both sides by

3

3 x = \frac{2 π}{3} + πn

Divide both sides by

3

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

Simplify

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

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

\frac{\frac{2 π}{3}}{3} + \frac{πn}{3} : \frac{2 π}{9} + \frac{πn}{3}

\frac{\frac{2 π}{3}}{3} + \frac{πn}{3}

\frac{\frac{2 π}{3}}{3} = \frac{2 π}{9}

\frac{\frac{2 π}{3}}{3}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{2 π}{3 \cdot 3}

Multiply the numbers:

3 \cdot 3 = 9

= \frac{2 π}{9}

= \frac{2 π}{9} + \frac{πn}{3}

x = \frac{2 π}{9} + \frac{πn}{3}

x = \frac{2 π}{9} + \frac{πn}{3}

x = \frac{2 π}{9} + \frac{πn}{3}

x = \frac{2 π}{9} + \frac{πn}{3}

Combine all the solutions

x = \frac{π}{9} + \frac{πn}{3}, x = \frac{2 π}{9} + \frac{πn}{3}

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Frequently Asked Questions (FAQ)

What is the general solution for 2tan^2(3x)=6 ?
The general solution for 2tan^2(3x)=6 is x= pi/9+(pin)/3 ,x=(2pi)/9+(pin)/3