What is the general solution for 6tan^4(x)-tan^2(x)-2=0 ?

The general solution for 6tan^4(x)-tan^2(x)-2=0 is x=0.68471…+pin,x=-0.68471…+pin

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6tan^4(x)-tan^2(x)-2=0

Solution

6 tan^{4} (x) - tan^{2} (x) - 2 = 0

Solution

x = 0.68471 \dots + πn, x = - 0.68471 \dots + πn

Degrees

x = 39.23152 \dots^{\circ} + 18 0^{\circ} n, x = - 39.23152 \dots^{\circ} + 18 0^{\circ} n

Solution steps

6 tan^{4} (x) - tan^{2} (x) - 2 = 0

Solve by substitution

6 tan^{4} (x) - tan^{2} (x) - 2 = 0

Let:

tan (x) = u

6 u^{4} - u^{2} - 2 = 0

6 u^{4} - u^{2} - 2 = 0 : u = \frac{2}{3}, u = - \frac{2}{3}, u = i \frac{1}{2}, u = - i \frac{1}{2}

6 u^{4} - u^{2} - 2 = 0

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

6 v^{2} - v - 2 = 0

Solve

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

6 v^{2} - v - 2 = 0

Solve with the quadratic formula

6 v^{2} - v - 2 = 0

Quadratic Equation Formula:

For

a = 6, b = - 1, c = - 2

v_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 6 ( - 2 )}{2 \cdot 6}

v_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 6 ( - 2 )}{2 \cdot 6}

(- 1)^{2} - 4 \cdot 6 (- 2) = 7

(- 1)^{2} - 4 \cdot 6 (- 2)

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 6 \cdot 2

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 6 \cdot 2 = 48

4 \cdot 6 \cdot 2

Multiply the numbers:

4 \cdot 6 \cdot 2 = 48

= 48

= 1 + 48

Add the numbers:

1 + 48 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

v_{1, 2} = \frac{- ( - 1 ) \pm 7}{2 \cdot 6}

Separate the solutions

v_{1} = \frac{- ( - 1 ) + 7}{2 \cdot 6}, v_{2} = \frac{- ( - 1 ) - 7}{2 \cdot 6}

v = \frac{- ( - 1 ) + 7}{2 \cdot 6} : \frac{2}{3}

\frac{- ( - 1 ) + 7}{2 \cdot 6}

Apply rule

- (- a) = a

= \frac{1 + 7}{2 \cdot 6}

Add the numbers:

1 + 7 = 8

= \frac{8}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{8}{12}

Cancel the common factor:

4

= \frac{2}{3}

v = \frac{- ( - 1 ) - 7}{2 \cdot 6} : - \frac{1}{2}

\frac{- ( - 1 ) - 7}{2 \cdot 6}

Apply rule

- (- a) = a

= \frac{1 - 7}{2 \cdot 6}

Subtract the numbers:

1 - 7 = - 6

= \frac{- 6}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 6}{12}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{6}{12}

Cancel the common factor:

6

= - \frac{1}{2}

The solutions to the quadratic equation are:

v = \frac{2}{3}, v = - \frac{1}{2}

v = \frac{2}{3}, v = - \frac{1}{2}

Substitute back

v = u^{2},

solve for

u

Solve

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

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

For

x^{2} = f (a)

the solutions are

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

u = \frac{2}{3}, u = - \frac{2}{3}

Solve

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

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

For

x^{2} = f (a)

the solutions are

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

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

Simplify

- \frac{1}{2} : i \frac{1}{2}

- \frac{1}{2}

Apply radical rule:

- a = - 1 a

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

= - 1 \frac{1}{2}

Apply imaginary number rule:

- 1 = i

= i \frac{1}{2}

Simplify

- - \frac{1}{2} : - i \frac{1}{2}

- - \frac{1}{2}

Simplify

- \frac{1}{2} : i \frac{1}{2}

- \frac{1}{2}

Apply radical rule:

- a = - 1 a

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

= - 1 \frac{1}{2}

Apply imaginary number rule:

- 1 = i

= i \frac{1}{2}

= - i \frac{1}{2}

u = i \frac{1}{2}, u = - i \frac{1}{2}

The solutions are

u = \frac{2}{3}, u = - \frac{2}{3}, u = i \frac{1}{2}, u = - i \frac{1}{2}

Substitute back

u = tan (x)

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

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

tan (x) = \frac{2}{3} : x = arctan (\frac{2}{3}) + πn

tan (x) = \frac{2}{3}

Apply trig inverse properties

tan (x) = \frac{2}{3}

General solutions for

tan (x) = \frac{2}{3}

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

x = arctan (\frac{2}{3}) + πn

x = arctan (\frac{2}{3}) + πn

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

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

Apply trig inverse properties

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

General solutions for

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

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

x = arctan (- \frac{2}{3}) + πn

x = arctan (- \frac{2}{3}) + πn

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

No Solution

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

No Solution

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

No Solution

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

No Solution

Combine all the solutions

x = arctan (\frac{2}{3}) + πn, x = arctan (- \frac{2}{3}) + πn

Show solutions in decimal form

x = 0.68471 \dots + πn, x = - 0.68471 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 6tan^4(x)-tan^2(x)-2=0 ?
The general solution for 6tan^4(x)-tan^2(x)-2=0 is x=0.68471…+pin,x=-0.68471…+pin