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

The general solution for tan(x)=2+tan(3x) is x= pi/4+pin,x=1.17809…+pin,x=-0.39269…+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Trigonometry >

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

Solution

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

Solution

x = \frac{π}{4} + πn, x = 1.17809 \dots + πn, x = - 0.39269 \dots + πn

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n, x = 67. 5^{\circ} + 18 0^{\circ} n, x = - 22. 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

2 + tan (3 x)

from both sides

tan (x) - 2 - tan (3 x) = 0

Rewrite using trig identities

- 2 - tan (3 x) + tan (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 )}

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

Combine the fractions

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

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

Convert element to fraction:

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

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

Expand

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

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

Simplify

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

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

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

1 \cdot tan (x)

Multiply:

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

= tan (x)

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

3 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)

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

Add the numbers:

2 + 1 = 3

= 3 tan^{3} (x)

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

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

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

Simplify

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

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

Add similar elements:

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

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

Add similar elements:

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

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

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

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

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

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

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

Solve by substitution

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

Let:

tan (x) = u

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

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

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

Multiply both sides by

1 - 3 u^{2}

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

Multiply both sides by

1 - 3 u^{2}

- 2 (1 - 3 u^{2}) + \frac{- 2 u - 2 u ^{3}}{1 - 3 u ^{2}} (1 - 3 u^{2}) = 0 \cdot (1 - 3 u^{2})

Simplify

- 2 (1 - 3 u^{2}) + \frac{- 2 u - 2 u ^{3}}{1 - 3 u ^{2}} (1 - 3 u^{2}) = 0 \cdot (1 - 3 u^{2})

Simplify

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

\frac{- 2 u - 2 u ^{3}}{1 - 3 u ^{2}} (1 - 3 u^{2})

Multiply fractions:

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

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

Cancel the common factor:

1 - 3 u^{2}

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

Simplify

0 \cdot (1 - 3 u^{2}) : 0

0 \cdot (1 - 3 u^{2})

Apply rule

0 \cdot a = 0

= 0

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

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

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

Solve

- 2 (1 - 3 u^{2}) - 2 u - 2 u^{3} = 0 : u = 1, u = 1 + 2, u = 1 - 2

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

Factor

- 2 (1 - 3 u^{2}) - 2 u - 2 u^{3} : - 2 (u - 1) (u^{2} - 2 u - 1)

- 2 (1 - 3 u^{2}) - 2 u - 2 u^{3}

Factor out common term

2

= - 2 (1 - u^{2} \cdot 3 + u + u^{3})

Factor

u^{3} - 3 u^{2} + u + 1 : (u - 1) (u^{2} - 2 u - 1)

1 - u^{2} \cdot 3 + u + u^{3}

Use the rational root theorem

a_{0} = 1, a_{n} = 1

The dividers of

a_{0} : 1,

The dividers of

a_{n} : 1

Therefore, check the following rational numbers:

\pm \frac{1}{1}

\frac{1}{1}

is a root of the expression, so factor out

u - 1

= (u - 1) \frac{u ^{3} - 3 u ^{2} + u + 1}{u - 1}

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

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

Divide

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

Divide the leading coefficients of the numerator

u^{3} - 3 u^{2} + u + 1

and the divisor

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

Quotient = u^{2}

Multiply

u - 1

u^{2} : u^{3} - u^{2}

Subtract

u^{3} - u^{2}

from

u^{3} - 3 u^{2} + u + 1

to get new remainder

Remainder = - 2 u^{2} + u + 1

Therefore

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

= u^{2} + \frac{- 2 u ^{2} + u + 1}{u - 1}

Divide

\frac{- 2 u ^{2} + u + 1}{u - 1} : \frac{- 2 u ^{2} + u + 1}{u - 1} = - 2 u + \frac{- u + 1}{u - 1}

Divide the leading coefficients of the numerator

- 2 u^{2} + u + 1

and the divisor

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

Quotient = - 2 u

Multiply

u - 1

- 2 u : - 2 u^{2} + 2 u

Subtract

- 2 u^{2} + 2 u

from

- 2 u^{2} + u + 1

to get new remainder

Remainder = - u + 1

Therefore

\frac{- 2 u ^{2} + u + 1}{u - 1} = - 2 u + \frac{- u + 1}{u - 1}

= u^{2} - 2 u + \frac{- u + 1}{u - 1}

Divide

\frac{- u + 1}{u - 1} : \frac{- u + 1}{u - 1} = - 1

Divide the leading coefficients of the numerator

- u + 1

and the divisor

u - 1 : \frac{- u}{u} = - 1

Quotient = - 1

Multiply

u - 1

- 1 : - u + 1

Subtract

- u + 1

from

- u + 1

to get new remainder

Remainder = 0

Therefore

\frac{- u + 1}{u - 1} = - 1

= u^{2} - 2 u - 1

= (u - 1) (u^{2} - 2 u - 1)

= - 2 (u - 1) (u^{2} - 2 u - 1)

- 2 (u - 1) (u^{2} - 2 u - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

u - 1 = 0 : u = 1

u - 1 = 0

Move

1

to the right side

u - 1 = 0

Add

1

to both sides

u - 1 + 1 = 0 + 1

Simplify

u = 1

u = 1

Solve

u^{2} - 2 u - 1 = 0 : u = 1 + 2, u = 1 - 2

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 1, b = - 2, c = - 1

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 1) = 22

(- 2)^{2} - 4 \cdot 1 \cdot (- 1)

Apply rule

- (- a) = a

= (- 2)^{2} + 4 \cdot 1 \cdot 1

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 2)^{2} = 2^{2}

= 2^{2} + 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 2^{2} + 4

2^{2} = 4

= 4 + 4

Add the numbers:

4 + 4 = 8

= 8

Prime factorization of

8 : 2^{3}

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

Apply exponent rule:

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

= 2^{2} \cdot 2

Apply radical rule:

= 2 2^{2}

Apply radical rule:

2^{2} = 2

= 22

u_{1, 2} = \frac{- ( - 2 ) \pm 2 2}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 2 2}{2 \cdot 1}, u_{2} = \frac{- ( - 2 ) - 2 2}{2 \cdot 1}

u = \frac{- ( - 2 ) + 2 2}{2 \cdot 1} : 1 + 2

\frac{- ( - 2 ) + 2 2}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{2 + 2 2}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2 + 2 2}{2}

Factor

2 + 22 : 2 (1 + 2)

2 + 22

Rewrite as

= 2 \cdot 1 + 22

Factor out common term

2

= 2 (1 + 2)

= \frac{2 ( 1 + 2 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= 1 + 2

u = \frac{- ( - 2 ) - 2 2}{2 \cdot 1} : 1 - 2

\frac{- ( - 2 ) - 2 2}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{2 - 2 2}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2 - 2 2}{2}

Factor

2 - 22 : 2 (1 - 2)

2 - 22

Rewrite as

= 2 \cdot 1 - 22

Factor out common term

2

= 2 (1 - 2)

= \frac{2 ( 1 - 2 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= 1 - 2

The solutions to the quadratic equation are:

u = 1 + 2, u = 1 - 2

The solutions are

u = 1, u = 1 + 2, u = 1 - 2

u = 1, u = 1 + 2, u = 1 - 2

Verify Solutions

Find undefined (singularity) points:

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

Take the denominator(s) of

- 2 + \frac{- 2 u - 2 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 = 1, u = 1 + 2, u = 1 - 2

Substitute back

u = tan (x)

tan (x) = 1, tan (x) = 1 + 2, tan (x) = 1 - 2

tan (x) = 1, tan (x) = 1 + 2, tan (x) = 1 - 2

tan (x) = 1 : x = \frac{π}{4} + πn

tan (x) = 1

General solutions for

tan (x) = 1

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}{4} + πn

x = \frac{π}{4} + πn

tan (x) = 1 + 2 : x = arctan (1 + 2) + πn

tan (x) = 1 + 2

Apply trig inverse properties

tan (x) = 1 + 2

General solutions for

tan (x) = 1 + 2

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

x = arctan (1 + 2) + πn

x = arctan (1 + 2) + πn

tan (x) = 1 - 2 : x = arctan (1 - 2) + πn

tan (x) = 1 - 2

Apply trig inverse properties

tan (x) = 1 - 2

General solutions for

tan (x) = 1 - 2

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

x = arctan (1 - 2) + πn

x = arctan (1 - 2) + πn

Combine all the solutions

x = \frac{π}{4} + πn, x = arctan (1 + 2) + πn, x = arctan (1 - 2) + πn

Show solutions in decimal form

x = \frac{π}{4} + πn, x = 1.17809 \dots + πn, x = - 0.39269 \dots + πn

Graph

Popular Examples

2sin(x)-3=0

sin^2(x)=2sin^2(x/2)

5sin(x)=cos(x)-4

-21tan(x)+3sqrt(3)=-3(sqrt(3)+tan(x))

tan(y)=-1

Frequently Asked Questions (FAQ)

What is the general solution for tan(x)=2+tan(3x) ?
The general solution for tan(x)=2+tan(3x) is x= pi/4+pin,x=1.17809…+pin,x=-0.39269…+pin