What is the general solution for tan^4(x)-2tan^2(x)-1=0 ?

The general solution for tan^4(x)-2tan^2(x)-1=0 is x=0.99893…+pin,x=-0.99893…+pin

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tan^4(x)-2tan^2(x)-1=0

Solution

tan^{4} (x) - 2 tan^{2} (x) - 1 = 0

Solution

x = 0.99893 \dots + πn, x = - 0.99893 \dots + πn

Degrees

x = 57.23490 \dots^{\circ} + 18 0^{\circ} n, x = - 57.23490 \dots^{\circ} + 18 0^{\circ} n

Solution steps

tan^{4} (x) - 2 tan^{2} (x) - 1 = 0

Solve by substitution

tan^{4} (x) - 2 tan^{2} (x) - 1 = 0

Let:

tan (x) = u

u^{4} - 2 u^{2} - 1 = 0

u^{4} - 2 u^{2} - 1 = 0 : u = 1 + 2, u = - 1 + 2, u = i - 1 + 2, u = - i - 1 + 2

u^{4} - 2 u^{2} - 1 = 0

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

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

Solve

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

v_{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

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

Separate the solutions

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

v = \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

v = \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:

v = 1 + 2, v = 1 - 2

v = 1 + 2, v = 1 - 2

Substitute back

v = u^{2},

solve for

u

Solve

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

u^{2} = 1 + 2

For

x^{2} = f (a)

the solutions are

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

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

Solve

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

u^{2} = 1 - 2

For

x^{2} = f (a)

the solutions are

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

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

Simplify

1 - 2 : i - 1 + 2

1 - 2

Apply imaginary number rule:

- a = i a

= i 2 - 1

Simplify

- 1 - 2 : - i - 1 + 2

- 1 - 2

Apply imaginary number rule:

- a = i a

= - i 2 - 1

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

The solutions are

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

Substitute back

u = tan (x)

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

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

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

tan (x) = i - 1 + 2 :

No Solution

tan (x) = i - 1 + 2

No Solution

tan (x) = - i - 1 + 2 :

No Solution

tan (x) = - i - 1 + 2

No Solution

Combine all the solutions

x = arctan (1 + 2) + πn, x = arctan (- 1 + 2) + πn

Show solutions in decimal form

x = 0.99893 \dots + πn, x = - 0.99893 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for tan^4(x)-2tan^2(x)-1=0 ?
The general solution for tan^4(x)-2tan^2(x)-1=0 is x=0.99893…+pin,x=-0.99893…+pin