What is the general solution for 2tan^4(x)-tan^2(x)-15=0 ?

The general solution for 2tan^4(x)-tan^2(x)-15=0 is x= pi/3+pin,x=(2pi)/3+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2tan^4(x)-tan^2(x)-15=0

Solution

2 tan^{4} (x) - tan^{2} (x) - 15 = 0

Solution

x = \frac{π}{3} + πn, x = \frac{2 π}{3} + πn

Degrees

x = 6 0^{\circ} + 18 0^{\circ} n, x = 12 0^{\circ} + 18 0^{\circ} n

Solution steps

2 tan^{4} (x) - tan^{2} (x) - 15 = 0

Solve by substitution

2 tan^{4} (x) - tan^{2} (x) - 15 = 0

Let:

tan (x) = u

2 u^{4} - u^{2} - 15 = 0

2 u^{4} - u^{2} - 15 = 0 : u = 3, u = - 3, u = i \frac{5}{2}, u = - i \frac{5}{2}

2 u^{4} - u^{2} - 15 = 0

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

2 v^{2} - v - 15 = 0

Solve

2 v^{2} - v - 15 = 0 : v = 3, v = - \frac{5}{2}

2 v^{2} - v - 15 = 0

Solve with the quadratic formula

2 v^{2} - v - 15 = 0

Quadratic Equation Formula:

For

a = 2, b = - 1, c = - 15

v_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 2 ( - 15 )}{2 \cdot 2}

v_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 2 ( - 15 )}{2 \cdot 2}

(- 1)^{2} - 4 \cdot 2 (- 15) = 11

(- 1)^{2} - 4 \cdot 2 (- 15)

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 2 \cdot 15

(- 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 2 \cdot 15 = 120

4 \cdot 2 \cdot 15

Multiply the numbers:

4 \cdot 2 \cdot 15 = 120

= 120

= 1 + 120

Add the numbers:

1 + 120 = 121

= 121

Factor the number:

121 = 1 1^{2}

= 1 1^{2}

Apply radical rule:

1 1^{2} = 11

= 11

v_{1, 2} = \frac{- ( - 1 ) \pm 11}{2 \cdot 2}

Separate the solutions

v_{1} = \frac{- ( - 1 ) + 11}{2 \cdot 2}, v_{2} = \frac{- ( - 1 ) - 11}{2 \cdot 2}

v = \frac{- ( - 1 ) + 11}{2 \cdot 2} : 3

\frac{- ( - 1 ) + 11}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{1 + 11}{2 \cdot 2}

Add the numbers:

1 + 11 = 12

= \frac{12}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{12}{4}

Divide the numbers:

\frac{12}{4} = 3

= 3

v = \frac{- ( - 1 ) - 11}{2 \cdot 2} : - \frac{5}{2}

\frac{- ( - 1 ) - 11}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{1 - 11}{2 \cdot 2}

Subtract the numbers:

1 - 11 = - 10

= \frac{- 10}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 10}{4}

Apply the fraction rule:

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

= - \frac{10}{4}

Cancel the common factor:

2

= - \frac{5}{2}

The solutions to the quadratic equation are:

v = 3, v = - \frac{5}{2}

v = 3, v = - \frac{5}{2}

Substitute back

v = u^{2},

solve for

u

Solve

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

u^{2} = 3

For

x^{2} = f (a)

the solutions are

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

u = 3, u = - 3

Solve

u^{2} = - \frac{5}{2} : u = i \frac{5}{2}, u = - i \frac{5}{2}

u^{2} = - \frac{5}{2}

For

x^{2} = f (a)

the solutions are

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

u = - \frac{5}{2}, u = - - \frac{5}{2}

Simplify

- \frac{5}{2} : i \frac{5}{2}

- \frac{5}{2}

Apply radical rule:

- a = - 1 a

- \frac{5}{2} = - 1 \frac{5}{2}

= - 1 \frac{5}{2}

Apply imaginary number rule:

- 1 = i

= i \frac{5}{2}

Simplify

- - \frac{5}{2} : - i \frac{5}{2}

- - \frac{5}{2}

Simplify

- \frac{5}{2} : i \frac{5}{2}

- \frac{5}{2}

Apply radical rule:

- a = - 1 a

- \frac{5}{2} = - 1 \frac{5}{2}

= - 1 \frac{5}{2}

Apply imaginary number rule:

- 1 = i

= i \frac{5}{2}

= - i \frac{5}{2}

u = i \frac{5}{2}, u = - i \frac{5}{2}

The solutions are

u = 3, u = - 3, u = i \frac{5}{2}, u = - i \frac{5}{2}

Substitute back

u = tan (x)

tan (x) = 3, tan (x) = - 3, tan (x) = i \frac{5}{2}, tan (x) = - i \frac{5}{2}

tan (x) = 3, tan (x) = - 3, tan (x) = i \frac{5}{2}, tan (x) = - i \frac{5}{2}

tan (x) = 3 : x = \frac{π}{3} + πn

tan (x) = 3

General solutions for

tan (x) = 3

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}{3} + πn

x = \frac{π}{3} + πn

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

tan (x) = - 3

General solutions for

tan (x) = - 3

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{2 π}{3} + πn

x = \frac{2 π}{3} + πn

tan (x) = i \frac{5}{2} :

No Solution

tan (x) = i \frac{5}{2}

No Solution

tan (x) = - i \frac{5}{2} :

No Solution

tan (x) = - i \frac{5}{2}

No Solution

Combine all the solutions

x = \frac{π}{3} + πn, x = \frac{2 π}{3} + πn

Popular Examples

cos(x)=-sqrt(1/2)

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

(30)/(sin(8))=(120)/(sin(x))

\frac{30}{sin ( 8 )} = \frac{120}{sin ( x )}

tan(x)+cot(x)-2=0

tan (x) + cot (x) - 2 = 0

4sin(A)-csc(A)=0

4 sin (A) - csc (A) = 0

6cos(θ)=-6(1+cos(θ))

6 cos (θ) = - 6 (1 + cos (θ))

Frequently Asked Questions (FAQ)

What is the general solution for 2tan^4(x)-tan^2(x)-15=0 ?
The general solution for 2tan^4(x)-tan^2(x)-15=0 is x= pi/3+pin,x=(2pi)/3+pin