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

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

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tan^4(x)-2=tan^2(x)+sec^2(x)

Solution

tan^{4} (x) - 2 = tan^{2} (x) + sec^{2} (x)

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

tan^{4} (x) - 2 = tan^{2} (x) + sec^{2} (x)

Subtract

tan^{2} (x) + sec^{2} (x)

from both sides

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

Rewrite using trig identities

- 2 - sec^{2} (x) - tan^{2} (x) + tan^{4} (x)

Use the Pythagorean identity:

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

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

Simplify

- 2 - (tan^{2} (x) + 1) - tan^{2} (x) + tan^{4} (x) : tan^{4} (x) - 2 tan^{2} (x) - 3

- 2 - (tan^{2} (x) + 1) - tan^{2} (x) + tan^{4} (x)

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

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

Distribute parentheses

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

Apply minus-plus rules

+ (- a) = - a

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

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

Simplify

- 2 - tan^{2} (x) - 1 - tan^{2} (x) + tan^{4} (x) : tan^{4} (x) - 2 tan^{2} (x) - 3

- 2 - tan^{2} (x) - 1 - tan^{2} (x) + tan^{4} (x)

Group like terms

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

Add similar elements:

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

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

Subtract the numbers:

- 2 - 1 = - 3

= tan^{4} (x) - 2 tan^{2} (x) - 3

= tan^{4} (x) - 2 tan^{2} (x) - 3

= tan^{4} (x) - 2 tan^{2} (x) - 3

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

Solve by substitution

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

Let:

tan (x) = u

- 3 + u^{4} - 2 u^{2} = 0

- 3 + u^{4} - 2 u^{2} = 0 : u = 3, u = - 3, u = i, u = - i

- 3 + u^{4} - 2 u^{2} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

u^{4} - 2 u^{2} - 3 = 0

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

v^{2} - 2 v - 3 = 0

Solve

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

v^{2} - 2 v - 3 = 0

Solve with the quadratic formula

v^{2} - 2 v - 3 = 0

Quadratic Equation Formula:

For

a = 1, b = - 2, c = - 3

v_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

v_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 3) = 4

(- 2)^{2} - 4 \cdot 1 \cdot (- 3)

Apply rule

- (- a) = a

= (- 2)^{2} + 4 \cdot 1 \cdot 3

Apply exponent rule:

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

n

is even

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

= 2^{2} + 4 \cdot 1 \cdot 3

Multiply the numbers:

4 \cdot 1 \cdot 3 = 12

= 2^{2} + 12

2^{2} = 4

= 4 + 12

Add the numbers:

4 + 12 = 16

= 16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

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

Separate the solutions

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

v = \frac{- ( - 2 ) + 4}{2 \cdot 1} : 3

\frac{- ( - 2 ) + 4}{2 \cdot 1}

Apply rule

- (- a) = a

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

Add the numbers:

2 + 4 = 6

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{6}{2}

Divide the numbers:

\frac{6}{2} = 3

= 3

v = \frac{- ( - 2 ) - 4}{2 \cdot 1} : - 1

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

Apply rule

- (- a) = a

= \frac{2 - 4}{2 \cdot 1}

Subtract the numbers:

2 - 4 = - 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{2}

Apply the fraction rule:

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

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

v = 3, v = - 1

v = 3, v = - 1

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} = - 1 : u = i, u = - i

u^{2} = - 1

For

x^{2} = f (a)

the solutions are

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

u = - 1, u = - - 1

Simplify

- 1 : i

- 1

Apply imaginary number rule:

- 1 = i

= i

Simplify

- - 1 : - i

- - 1

Apply imaginary number rule:

- 1 = i

= - i

u = i, u = - i

The solutions are

u = 3, u = - 3, u = i, u = - i

Substitute back

u = tan (x)

tan (x) = 3, tan (x) = - 3, tan (x) = i, tan (x) = - i

tan (x) = 3, tan (x) = - 3, tan (x) = i, tan (x) = - i

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 :

No Solution

tan (x) = i

No Solution

tan (x) = - i :

No Solution

tan (x) = - i

No Solution

Combine all the solutions

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

Popular Examples

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cos (x - \frac{π}{4}) = - sin (x)

sin(x)= 5/13 ,cos(x)

sin (x) = \frac{5}{13}, cos (x)

tan(θ)=(11.9)/(10)

tan (θ) = \frac{11.9}{10}

tanh(x)= 3/5

tanh (x) = \frac{3}{5}

csc(θ)+2.402=0

csc (θ) + 2.402 = 0

Frequently Asked Questions (FAQ)

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