What is the general solution for 7tan(3x)-21tan(x)=0 ?

The general solution for 7tan(3x)-21tan(x)=0 is x=pin

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Popular Trigonometry >

7tan(3x)-21tan(x)=0

Solution

7 tan (3 x) - 21 tan (x) = 0

Solution

x = πn

Degrees

x = 0^{\circ} + 18 0^{\circ} n

Solution steps

7 tan (3 x) - 21 tan (x) = 0

Rewrite using trig identities

- 21 tan (x) + 7 tan (3 x)

tan (3 x) = \frac{3 tan ( x ) - tan ^{3} ( x )}{1 - 3 tan ^{2} ( x )}

tan (3 x)

Rewrite using trig identities

tan (3 x)

Rewrite as

= tan (2 x + x)

Use the Angle Sum identity:

tan (s + t) = \frac{tan ( s ) + tan ( t )}{1 - tan ( s ) tan ( t )}

= \frac{tan ( 2 x ) + tan ( x )}{1 - tan ( 2 x ) tan ( x )}

= \frac{tan ( 2 x ) + tan ( x )}{1 - tan ( 2 x ) tan ( x )}

Use the Double Angle identity:

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

= \frac{\frac{2 t a n ( x )}{1 - t a n ^{2} ( x )} + tan ( x )}{1 - \frac{2 t a n ( x )}{1 - t a n ^{2} ( x )} tan ( x )}

Simplify

\frac{\frac{2 t a n ( x )}{1 - t a n ^{2} ( x )} + tan ( x )}{1 - \frac{2 t a n ( x )}{1 - t a n ^{2} ( x )} tan ( x )} : \frac{3 tan ( x ) - tan ^{3} ( x )}{1 - 3 tan ^{2} ( x )}

\frac{\frac{2 t a n ( x )}{1 - t a n ^{2} ( x )} + tan ( x )}{1 - \frac{2 t a n ( x )}{1 - t a n ^{2} ( x )} tan ( x )}

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

\frac{2 tan ( x )}{1 - tan ^{2} ( x )} tan (x)

Multiply fractions:

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

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

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

2 tan (x) tan (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

tan (x) tan (x) = tan^{1 + 1} (x)

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

Add the numbers:

1 + 1 = 2

= 2 tan^{2} (x)

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

= \frac{\frac{2 t a n ( x )}{- t a n ^{2} ( x ) + 1} + tan ( x )}{1 - \frac{2 t a n ^{2} ( x )}{- t a n ^{2} ( x ) + 1}}

Join

\frac{2 tan ( x )}{1 - tan ^{2} ( x )} + tan (x) : \frac{3 tan ( x ) - tan ^{3} ( x )}{1 - tan ^{2} ( x )}

\frac{2 tan ( x )}{1 - tan ^{2} ( x )} + tan (x)

Convert element to fraction:

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

= \frac{2 tan ( x )}{1 - tan ^{2} ( x )} + \frac{tan ( x ) ( 1 - tan ^{2} ( x ) )}{1 - tan ^{2} ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{2 tan ( x ) + tan ( x ) ( 1 - tan ^{2} ( x ) )}{1 - tan ^{2} ( x )}

Expand

2 tan (x) + tan (x) (1 - tan^{2} (x)) : 3 tan (x) - tan^{3} (x)

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

Expand

tan (x) (1 - tan^{2} (x)) : tan (x) - tan^{3} (x)

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

Apply the distributive law:

a (b - c) = ab - a c

a = tan (x), b = 1, c = tan^{2} (x)

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

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

Simplify

1 \cdot tan (x) - tan^{2} (x) tan (x) : tan (x) - tan^{3} (x)

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

1 \cdot tan (x) = tan (x)

1 tan (x)

Multiply:

1 \cdot tan (x) = tan (x)

= tan (x)

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

tan^{2} (x) tan (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= tan^{2 + 1} (x)

Add the numbers:

2 + 1 = 3

= tan^{3} (x)

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

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

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

Add similar elements:

2 tan (x) + tan (x) = 3 tan (x)

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

= \frac{3 tan ( x ) - tan ^{3} ( x )}{1 - tan ^{2} ( x )}

= \frac{\frac{3 t a n ( x ) - t a n ^{3} ( x )}{1 - t a n ^{2} ( x )}}{1 - \frac{2 t a n ^{2} ( x )}{- t a n ^{2} ( x ) + 1}}

Apply the fraction rule:

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

= \frac{3 tan ( x ) - tan ^{3} ( x )}{( 1 - tan ^{2} ( x ) ) ( 1 - \frac{2 t a n ^{2} ( x )}{1 - t a n ^{2} ( x )} )}

Join

1 - \frac{2 tan ^{2} ( x )}{1 - tan ^{2} ( x )} : \frac{1 - 3 tan ^{2} ( x )}{1 - tan ^{2} ( x )}

1 - \frac{2 tan ^{2} ( x )}{1 - tan ^{2} ( x )}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

1 \cdot (1 - tan^{2} (x)) - 2 tan^{2} (x) = 1 - 3 tan^{2} (x)

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

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

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

Multiply:

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

= 1 - tan^{2} (x)

Remove parentheses:

(a) = a

= 1 - tan^{2} (x)

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

Add similar elements:

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

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

= \frac{1 - 3 tan ^{2} ( x )}{1 - tan ^{2} ( x )}

= \frac{3 tan ( x ) - tan ^{3} ( x )}{\frac{- 3 t a n ^{2} ( x ) + 1}{- t a n ^{2} ( x ) + 1} ( - tan ^{2} ( x ) + 1 )}

Multiply

(1 - tan^{2} (x)) \frac{1 - 3 tan ^{2} ( x )}{1 - tan ^{2} ( x )} : 1 - 3 tan^{2} (x)

(1 - tan^{2} (x)) \frac{1 - 3 tan ^{2} ( x )}{1 - tan ^{2} ( x )}

Multiply fractions:

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

= \frac{( 1 - 3 tan ^{2} ( x ) ) ( 1 - tan ^{2} ( x ) )}{1 - tan ^{2} ( x )}

Cancel the common factor:

1 - tan^{2} (x)

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

= \frac{3 tan ( x ) - tan ^{3} ( x )}{1 - 3 tan ^{2} ( x )}

= \frac{3 tan ( x ) - tan ^{3} ( x )}{1 - 3 tan ^{2} ( x )}

= - 21 tan (x) + 7 \cdot \frac{3 tan ( x ) - tan ^{3} ( x )}{1 - 3 tan ^{2} ( x )}

Simplify

- 21 tan (x) + 7 \cdot \frac{3 tan ( x ) - tan ^{3} ( x )}{1 - 3 tan ^{2} ( x )} : \frac{56 tan ^{3} ( x )}{1 - 3 tan ^{2} ( x )}

- 21 tan (x) + 7 \cdot \frac{3 tan ( x ) - tan ^{3} ( x )}{1 - 3 tan ^{2} ( x )}

Multiply

7 \cdot \frac{3 tan ( x ) - tan ^{3} ( x )}{1 - 3 tan ^{2} ( x )} : \frac{7 ( 3 tan ( x ) - tan ^{3} ( x ) )}{1 - 3 tan ^{2} ( x )}

7 \cdot \frac{3 tan ( x ) - tan ^{3} ( x )}{1 - 3 tan ^{2} ( x )}

Multiply fractions:

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

= \frac{( 3 tan ( x ) - tan ^{3} ( x ) ) \cdot 7}{1 - 3 tan ^{2} ( x )}

= - 21 tan (x) + \frac{7 ( 3 tan ( x ) - tan ^{3} ( x ) )}{- 3 tan ^{2} ( x ) + 1}

Convert element to fraction:

21 tan (x) = \frac{21 tan ( x ) ( 1 - 3 tan ^{2} ( x ) )}{1 - 3 tan ^{2} ( x )}

= \frac{( 3 tan ( x ) - tan ^{3} ( x ) ) \cdot 7}{1 - 3 tan ^{2} ( x )} - \frac{21 tan ( x ) ( 1 - 3 tan ^{2} ( x ) )}{1 - 3 tan ^{2} ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{( 3 tan ( x ) - tan ^{3} ( x ) ) \cdot 7 - 21 tan ( x ) ( 1 - 3 tan ^{2} ( x ) )}{1 - 3 tan ^{2} ( x )}

Expand

(3 tan (x) - tan^{3} (x)) \cdot 7 - 21 tan (x) (1 - 3 tan^{2} (x)) : 56 tan^{3} (x)

(3 tan (x) - tan^{3} (x)) \cdot 7 - 21 tan (x) (1 - 3 tan^{2} (x))

= 7 (3 tan (x) - tan^{3} (x)) - 21 tan (x) (1 - 3 tan^{2} (x))

Expand

7 (3 tan (x) - tan^{3} (x)) : 21 tan (x) - 7 tan^{3} (x)

7 (3 tan (x) - tan^{3} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 7, b = 3 tan (x), c = tan^{3} (x)

= 7 \cdot 3 tan (x) - 7 tan^{3} (x)

Multiply the numbers:

7 \cdot 3 = 21

= 21 tan (x) - 7 tan^{3} (x)

= 21 tan (x) - 7 tan^{3} (x) - 21 tan (x) (1 - 3 tan^{2} (x))

Expand

- 21 tan (x) (1 - 3 tan^{2} (x)) : - 21 tan (x) + 63 tan^{3} (x)

- 21 tan (x) (1 - 3 tan^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = - 21 tan (x), b = 1, c = 3 tan^{2} (x)

= - 21 tan (x) \cdot 1 - (- 21 tan (x)) \cdot 3 tan^{2} (x)

Apply minus-plus rules

- (- a) = a

= - 21 \cdot 1 \cdot tan (x) + 21 \cdot 3 tan^{2} (x) tan (x)

Simplify

- 21 \cdot 1 \cdot tan (x) + 21 \cdot 3 tan^{2} (x) tan (x) : - 21 tan (x) + 63 tan^{3} (x)

- 21 \cdot 1 \cdot tan (x) + 21 \cdot 3 tan^{2} (x) tan (x)

21 \cdot 1 \cdot tan (x) = 21 tan (x)

21 \cdot 1 \cdot tan (x)

Multiply the numbers:

21 \cdot 1 = 21

= 21 tan (x)

21 \cdot 3 tan^{2} (x) tan (x) = 63 tan^{3} (x)

21 \cdot 3 tan^{2} (x) tan (x)

Multiply the numbers:

21 \cdot 3 = 63

= 63 tan^{2} (x) tan (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= 63 tan^{2 + 1} (x)

Add the numbers:

2 + 1 = 3

= 63 tan^{3} (x)

= - 21 tan (x) + 63 tan^{3} (x)

= - 21 tan (x) + 63 tan^{3} (x)

= 21 tan (x) - 7 tan^{3} (x) - 21 tan (x) + 63 tan^{3} (x)

Simplify

21 tan (x) - 7 tan^{3} (x) - 21 tan (x) + 63 tan^{3} (x) : 56 tan^{3} (x)

21 tan (x) - 7 tan^{3} (x) - 21 tan (x) + 63 tan^{3} (x)

Add similar elements:

- 7 tan^{3} (x) + 63 tan^{3} (x) = 56 tan^{3} (x)

= 21 tan (x) + 56 tan^{3} (x) - 21 tan (x)

Add similar elements:

21 tan (x) - 21 tan (x) = 0

= 56 tan^{3} (x)

= 56 tan^{3} (x)

= \frac{56 tan ^{3} ( x )}{1 - 3 tan ^{2} ( x )}

= \frac{56 tan ^{3} ( x )}{1 - 3 tan ^{2} ( x )}

\frac{56 tan ^{3} ( x )}{1 - 3 tan ^{2} ( x )} = 0

Solve by substitution

\frac{56 tan ^{3} ( x )}{1 - 3 tan ^{2} ( x )} = 0

Let:

tan (x) = u

\frac{56 u ^{3}}{1 - 3 u ^{2}} = 0

\frac{56 u ^{3}}{1 - 3 u ^{2}} = 0 : u = 0

\frac{56 u ^{3}}{1 - 3 u ^{2}} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

56 u^{3} = 0

Solve

56 u^{3} = 0 : u = 0

56 u^{3} = 0

Divide both sides by

56

56 u^{3} = 0

Divide both sides by

56

56 u^{3} = 0

Divide both sides by

56

\frac{56 u ^{3}}{56} = \frac{0}{56}

Simplify

u^{3} = 0

u^{3} = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

u = 0

u = 0

Verify Solutions

Find undefined (singularity) points:

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

Take the denominator(s) of

\frac{56 u ^{3}}{1 - 3 u ^{2}}

and compare to zero

Solve

1 - 3 u^{2} = 0 : u = \frac{1}{3}, u = - \frac{1}{3}

1 - 3 u^{2} = 0

Move

1

to the right side

1 - 3 u^{2} = 0

Subtract

1

from both sides

1 - 3 u^{2} - 1 = 0 - 1

Simplify

- 3 u^{2} = - 1

- 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}

\frac{1}{3} = \frac{1}{3}

\frac{1}{3}

Apply radical rule:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= \frac{1}{3}

Apply radical rule:

1 = 1

1 = 1

= \frac{1}{3}

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

- \frac{1}{3}

Apply radical rule:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= - \frac{1}{3}

Apply radical rule:

1 = 1

1 = 1

= - \frac{1}{3}

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

The following points are undefined

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

Combine undefined points with solutions:

u = 0

Substitute back

u = tan (x)

tan (x) = 0

tan (x) = 0

tan (x) = 0 : x = πn

tan (x) = 0

General solutions for

tan (x) = 0

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 = 0 + πn

x = 0 + πn

Solve

x = 0 + πn : x = πn

x = 0 + πn

0 + πn = πn

x = πn

x = πn

Combine all the solutions

x = πn

Graph

Popular Examples

tan((3θ)/4)=(-sqrt(3))/3

Frequently Asked Questions (FAQ)

What is the general solution for 7tan(3x)-21tan(x)=0 ?
The general solution for 7tan(3x)-21tan(x)=0 is x=pin