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Popular >

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

Solution

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

Solution

x = \frac{π}{3} + πn, x = \frac{2 π}{3} + πn, x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

Degrees

x = 6 0^{\circ} + 18 0^{\circ} n, x = 12 0^{\circ} + 18 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

tan (x) = u

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

u^{4} + 1 = 4 u^{2} - 2 : u = 3, u = - 3, u = 1, u = - 1

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

Move

2

to the left side

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

Add

2

to both sides

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

Simplify

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

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

Move

4 u^{2}

to the left side

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

Subtract

4 u^{2}

from both sides

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

Simplify

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

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

Write in the standard form

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

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

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

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

Solve

v^{2} - 4 v + 3 = 0 : v = 3, v = 1

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 1, b = - 4, c = 3

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

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

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

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

Apply exponent rule:

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

n

is even

(- 4)^{2} = 4^{2}

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

Multiply the numbers:

4 \cdot 1 \cdot 3 = 12

= 4^{2} - 12

4^{2} = 16

= 16 - 12

Subtract the numbers:

16 - 12 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

4 - 2 = 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \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 = 1, u = - 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

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

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

The solutions are

u = 3, u = - 3, u = 1, u = - 1

Substitute back

u = tan (x)

tan (x) = 3, tan (x) = - 3, tan (x) = 1, tan (x) = - 1

tan (x) = 3, tan (x) = - 3, tan (x) = 1, tan (x) = - 1

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) = 1 : x = \frac{π}{4} + πn

tan (x) = 1

General solutions for

tan (x) = 1

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}{4} + πn

x = \frac{π}{4} + πn

tan (x) = - 1 : x = \frac{3 π}{4} + πn

tan (x) = - 1

General solutions for

tan (x) = - 1

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{3 π}{4} + πn

x = \frac{3 π}{4} + πn

Combine all the solutions

x = \frac{π}{3} + πn, x = \frac{2 π}{3} + πn, x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

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