What is the general solution for 3tan^2(3x)=1 ?

The general solution for 3tan^2(3x)=1 is x= pi/(18)+(pin)/3 ,x=-pi/(18)+(pin)/3

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3tan^2(3x)=1

Solution

3 tan^{2} (3 x) = 1

Solution

x = \frac{π}{18} + \frac{πn}{3}, x = - \frac{π}{18} + \frac{πn}{3}

Degrees

x = 1 0^{\circ} + 6 0^{\circ} n, x = - 1 0^{\circ} + 6 0^{\circ} n

Solution steps

3 tan^{2} (3 x) = 1

Solve by substitution

3 tan^{2} (3 x) = 1

Let:

tan (3 x) = u

3 u^{2} = 1

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

3 u^{2} = 1

Divide both sides by

3

3 u^{2} = 1

Divide both sides by

3

\frac{3 u ^{2}}{3} = \frac{1}{3}

Simplify

u^{2} = \frac{1}{3}

u^{2} = \frac{1}{3}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = \frac{1}{3}, u = - \frac{1}{3}

Substitute back

u = tan (3 x)

tan (3 x) = \frac{1}{3}, tan (3 x) = - \frac{1}{3}

tan (3 x) = \frac{1}{3}, tan (3 x) = - \frac{1}{3}

tan (3 x) = \frac{1}{3} : x = \frac{π}{18} + \frac{πn}{3}

tan (3 x) = \frac{1}{3}

Apply trig inverse properties

tan (3 x) = \frac{1}{3}

General solutions for

tan (3 x) = \frac{1}{3}

tan (x) = a \Rightarrow x = arctan (a) + πn

3 x = arctan (\frac{1}{3}) + πn

3 x = arctan (\frac{1}{3}) + πn

Solve

3 x = arctan (\frac{1}{3}) + πn : x = \frac{π}{18} + \frac{πn}{3}

3 x = arctan (\frac{1}{3}) + πn

Simplify

arctan (\frac{1}{3}) + πn : \frac{π}{6} + πn

arctan (\frac{1}{3}) + πn

Use the following trivial identity:

arctan (\frac{1}{3}) = \frac{π}{6}

x 0 \frac{3}{3} 13 arctan (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} arctan (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ}

= \frac{π}{6} + πn

3 x = \frac{π}{6} + πn

Divide both sides by

3

3 x = \frac{π}{6} + πn

Divide both sides by

3

\frac{3 x}{3} = \frac{\frac{π}{6}}{3} + \frac{πn}{3}

Simplify

\frac{3 x}{3} = \frac{\frac{π}{6}}{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}{6}}{3} + \frac{πn}{3} : \frac{π}{18} + \frac{πn}{3}

\frac{\frac{π}{6}}{3} + \frac{πn}{3}

\frac{\frac{π}{6}}{3} = \frac{π}{18}

\frac{\frac{π}{6}}{3}

Apply the fraction rule:

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

= \frac{π}{6 \cdot 3}

Multiply the numbers:

6 \cdot 3 = 18

= \frac{π}{18}

= \frac{π}{18} + \frac{πn}{3}

x = \frac{π}{18} + \frac{πn}{3}

x = \frac{π}{18} + \frac{πn}{3}

x = \frac{π}{18} + \frac{πn}{3}

x = \frac{π}{18} + \frac{πn}{3}

tan (3 x) = - \frac{1}{3} : x = - \frac{π}{18} + \frac{πn}{3}

tan (3 x) = - \frac{1}{3}

Apply trig inverse properties

tan (3 x) = - \frac{1}{3}

General solutions for

tan (3 x) = - \frac{1}{3}

tan (x) = - a \Rightarrow x = arctan (- a) + πn

3 x = arctan (- \frac{1}{3}) + πn

3 x = arctan (- \frac{1}{3}) + πn

Solve

3 x = arctan (- \frac{1}{3}) + πn : x = - \frac{π}{18} + \frac{πn}{3}

3 x = arctan (- \frac{1}{3}) + πn

Simplify

arctan (- \frac{1}{3}) + πn : - \frac{π}{6} + πn

arctan (- \frac{1}{3}) + πn

arctan (- \frac{1}{3}) = - \frac{π}{6}

arctan (- \frac{1}{3})

Use the following property:

arctan (- x) = - arctan (x)

arctan (- \frac{1}{3}) = - arctan (\frac{1}{3})

= - arctan (\frac{1}{3})

Use the following trivial identity:

arctan (\frac{1}{3}) = \frac{π}{6}

arctan (\frac{1}{3})

x 0 \frac{3}{3} 13 arctan (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} arctan (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ}

= \frac{π}{6}

= - \frac{π}{6}

= - \frac{π}{6} + πn

3 x = - \frac{π}{6} + πn

Divide both sides by

3

3 x = - \frac{π}{6} + πn

Divide both sides by

3

\frac{3 x}{3} = - \frac{\frac{π}{6}}{3} + \frac{πn}{3}

Simplify

\frac{3 x}{3} = - \frac{\frac{π}{6}}{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}{6}}{3} + \frac{πn}{3} : - \frac{π}{18} + \frac{πn}{3}

- \frac{\frac{π}{6}}{3} + \frac{πn}{3}

\frac{\frac{π}{6}}{3} = \frac{π}{18}

\frac{\frac{π}{6}}{3}

Apply the fraction rule:

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

= \frac{π}{6 \cdot 3}

Multiply the numbers:

6 \cdot 3 = 18

= \frac{π}{18}

= - \frac{π}{18} + \frac{πn}{3}

x = - \frac{π}{18} + \frac{πn}{3}

x = - \frac{π}{18} + \frac{πn}{3}

x = - \frac{π}{18} + \frac{πn}{3}

x = - \frac{π}{18} + \frac{πn}{3}

Combine all the solutions

x = \frac{π}{18} + \frac{πn}{3}, x = - \frac{π}{18} + \frac{πn}{3}

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Frequently Asked Questions (FAQ)

What is the general solution for 3tan^2(3x)=1 ?
The general solution for 3tan^2(3x)=1 is x= pi/(18)+(pin)/3 ,x=-pi/(18)+(pin)/3