What is the general solution for tan^2(x)+1/6+(tan(1))/3 =0 ?

The general solution for tan^2(x)+1/6+(tan(1))/3 =0 is No Solution for x\in\mathbb{R}

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tan^2(x)+1/6+(tan(1))/3 =0

Solution

tan^{2} (x) + \frac{1}{6} + \frac{tan ( 1 )}{3} = 0

Solution

No Solution for x \in R

Solution steps

tan^{2} (x) + \frac{1}{6} + \frac{tan ( 1 )}{3} = 0

Solve by substitution

tan^{2} (x) + \frac{1}{6} + \frac{tan ( 1 )}{3} = 0

Let:

tan (x) = u

u^{2} + \frac{1}{6} + \frac{tan ( 1 )}{3} = 0

u^{2} + \frac{1}{6} + \frac{tan ( 1 )}{3} = 0 : u = i \frac{1 + 2 tan ( 1 )}{6}, u = - i \frac{1 + 2 tan ( 1 )}{6}

u^{2} + \frac{1}{6} + \frac{tan ( 1 )}{3} = 0

Move

\frac{1}{6}

to the right side

u^{2} + \frac{1}{6} + \frac{tan ( 1 )}{3} = 0

Subtract

\frac{1}{6}

from both sides

u^{2} + \frac{1}{6} + \frac{tan ( 1 )}{3} - \frac{1}{6} = 0 - \frac{1}{6}

Simplify

u^{2} + \frac{tan ( 1 )}{3} = - \frac{1}{6}

u^{2} + \frac{tan ( 1 )}{3} = - \frac{1}{6}

Move

\frac{tan ( 1 )}{3}

to the right side

u^{2} + \frac{tan ( 1 )}{3} = - \frac{1}{6}

Subtract

\frac{tan ( 1 )}{3}

from both sides

u^{2} + \frac{tan ( 1 )}{3} - \frac{tan ( 1 )}{3} = - \frac{1}{6} - \frac{tan ( 1 )}{3}

Simplify

u^{2} = - \frac{1}{6} - \frac{tan ( 1 )}{3}

u^{2} = - \frac{1}{6} - \frac{tan ( 1 )}{3}

For

x^{2} = f (a)

the solutions are

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

u = - \frac{1}{6} - \frac{tan ( 1 )}{3}, u = - - \frac{1}{6} - \frac{tan ( 1 )}{3}

Simplify

- \frac{1}{6} - \frac{tan ( 1 )}{3} : i \frac{1 + 2 tan ( 1 )}{6}

- \frac{1}{6} - \frac{tan ( 1 )}{3}

Apply imaginary number rule:

- a = i a

= i \frac{1}{6} + \frac{tan ( 1 )}{3}

Rewrite

i \frac{1}{6} + \frac{tan ( 1 )}{3}

in standard complex form:

\frac{1 + 2 tan ( 1 )}{6} i

i \frac{1}{6} + \frac{tan ( 1 )}{3}

\frac{1}{6} + \frac{tan ( 1 )}{3} = \frac{1 + 2 tan ( 1 )}{6}

\frac{1}{6} + \frac{tan ( 1 )}{3}

Join

\frac{1}{6} + \frac{tan ( 1 )}{3} : \frac{1 + 2 tan ( 1 )}{6}

\frac{1}{6} + \frac{tan ( 1 )}{3}

Least Common Multiplier of

6, 3 : 6

6, 3

Least Common Multiplier (LCM)

Prime factorization of

6 : 2 \cdot 3

6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 3

Prime factorization of

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

Multiply each factor the greatest number of times it occurs in either

6

3

= 2 \cdot 3

Multiply the numbers:

2 \cdot 3 = 6

= 6

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

6

For

\frac{tan ( 1 )}{3} :

multiply the denominator and numerator by

2

\frac{tan ( 1 )}{3} = \frac{tan ( 1 ) \cdot 2}{3 \cdot 2} = \frac{tan ( 1 ) \cdot 2}{6}

= \frac{1}{6} + \frac{tan ( 1 ) \cdot 2}{6}

Since the denominators are equal, combine the fractions:

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

= \frac{1 + tan ( 1 ) \cdot 2}{6}

= \frac{1 + tan ( 1 ) \cdot 2}{6}

= \frac{1 + 2 tan ( 1 )}{6} i

= \frac{1 + 2 tan ( 1 )}{6} i

Simplify

- - \frac{1}{6} - \frac{tan ( 1 )}{3} : - i \frac{1 + 2 tan ( 1 )}{6}

- - \frac{1}{6} - \frac{tan ( 1 )}{3}

Apply imaginary number rule:

- a = i a

= - i \frac{1}{6} + \frac{tan ( 1 )}{3}

Rewrite

- i \frac{1}{6} + \frac{tan ( 1 )}{3}

in standard complex form:

- \frac{1 + 2 tan ( 1 )}{6} i

- i \frac{1}{6} + \frac{tan ( 1 )}{3}

- \frac{1}{6} + \frac{tan ( 1 )}{3} = - \frac{1 + 2 tan ( 1 )}{6}

- \frac{1}{6} + \frac{tan ( 1 )}{3}

Join

\frac{1}{6} + \frac{tan ( 1 )}{3} : \frac{1 + 2 tan ( 1 )}{6}

\frac{1}{6} + \frac{tan ( 1 )}{3}

Least Common Multiplier of

6, 3 : 6

6, 3

Least Common Multiplier (LCM)

Prime factorization of

6 : 2 \cdot 3

6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 3

Prime factorization of

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

Multiply each factor the greatest number of times it occurs in either

6

3

= 2 \cdot 3

Multiply the numbers:

2 \cdot 3 = 6

= 6

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

6

For

\frac{tan ( 1 )}{3} :

multiply the denominator and numerator by

2

\frac{tan ( 1 )}{3} = \frac{tan ( 1 ) \cdot 2}{3 \cdot 2} = \frac{tan ( 1 ) \cdot 2}{6}

= \frac{1}{6} + \frac{tan ( 1 ) \cdot 2}{6}

Since the denominators are equal, combine the fractions:

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

= \frac{1 + tan ( 1 ) \cdot 2}{6}

= - \frac{2 tan ( 1 ) + 1}{6}

= - \frac{1 + 2 tan ( 1 )}{6} i

= - \frac{1 + 2 tan ( 1 )}{6} i

u = i \frac{1 + 2 tan ( 1 )}{6}, u = - i \frac{1 + 2 tan ( 1 )}{6}

Substitute back

u = tan (x)

tan (x) = i \frac{1 + 2 tan ( 1 )}{6}, tan (x) = - i \frac{1 + 2 tan ( 1 )}{6}

tan (x) = i \frac{1 + 2 tan ( 1 )}{6}, tan (x) = - i \frac{1 + 2 tan ( 1 )}{6}

tan (x) = i \frac{1 + 2 tan ( 1 )}{6} :

No Solution

tan (x) = i \frac{1 + 2 tan ( 1 )}{6}

No Solution

tan (x) = - i \frac{1 + 2 tan ( 1 )}{6} :

No Solution

tan (x) = - i \frac{1 + 2 tan ( 1 )}{6}

No Solution

Combine all the solutions

No Solution for x \in R

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Frequently Asked Questions (FAQ)

What is the general solution for tan^2(x)+1/6+(tan(1))/3 =0 ?
The general solution for tan^2(x)+1/6+(tan(1))/3 =0 is No Solution for x\in\mathbb{R}