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Popular >

5tan^4(x)-10tan^2(x)+1=0

Solution

5 tan^{4} (x) - 10 tan^{2} (x) + 1 = 0

Solution

x = 0.94247 \dots + πn, x = - 0.94247 \dots + πn, x = 0.31415 \dots + πn, x = - 0.31415 \dots + πn

Degrees

x = 5 4^{\circ} + 18 0^{\circ} n, x = - 5 4^{\circ} + 18 0^{\circ} n, x = 1 8^{\circ} + 18 0^{\circ} n, x = - 1 8^{\circ} + 18 0^{\circ} n

Solution steps

5 tan^{4} (x) - 10 tan^{2} (x) + 1 = 0

Solve by substitution

5 tan^{4} (x) - 10 tan^{2} (x) + 1 = 0

Let:

tan (x) = u

5 u^{4} - 10 u^{2} + 1 = 0

5 u^{4} - 10 u^{2} + 1 = 0 : u = \frac{5 + 2 5}{5}, u = - \frac{5 + 2 5}{5}, u = \frac{5 - 2 5}{5}, u = - \frac{5 - 2 5}{5}

5 u^{4} - 10 u^{2} + 1 = 0

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

5 v^{2} - 10 v + 1 = 0

Solve

5 v^{2} - 10 v + 1 = 0 : v = \frac{5 + 2 5}{5}, v = \frac{5 - 2 5}{5}

5 v^{2} - 10 v + 1 = 0

Solve with the quadratic formula

5 v^{2} - 10 v + 1 = 0

Quadratic Equation Formula:

For

a = 5, b = - 10, c = 1

v_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 5 \cdot 1}{2 \cdot 5}

v_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 5 \cdot 1}{2 \cdot 5}

(- 10)^{2} - 4 \cdot 5 \cdot 1 = 45

(- 10)^{2} - 4 \cdot 5 \cdot 1

Apply exponent rule:

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

n

is even

(- 10)^{2} = 1 0^{2}

= 1 0^{2} - 4 \cdot 5 \cdot 1

Multiply the numbers:

4 \cdot 5 \cdot 1 = 20

= 1 0^{2} - 20

1 0^{2} = 100

= 100 - 20

Subtract the numbers:

100 - 20 = 80

= 80

Prime factorization of

80 : 2^{4} \cdot 5

80

80

divides by

280 = 40 \cdot 2

= 2 \cdot 40

40

divides by

240 = 20 \cdot 2

= 2 \cdot 2 \cdot 20

20

divides by

220 = 10 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 10

10

divides by

210 = 5 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 5

2, 5

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 5

= 2^{4} \cdot 5

= 2^{4} \cdot 5

Apply radical rule:

= 5 2^{4}

Apply radical rule:

2^{4} = 2^{\frac{4}{2}} = 2^{2}

= 2^{2} 5

Refine

= 45

v_{1, 2} = \frac{- ( - 10 ) \pm 4 5}{2 \cdot 5}

Separate the solutions

v_{1} = \frac{- ( - 10 ) + 4 5}{2 \cdot 5}, v_{2} = \frac{- ( - 10 ) - 4 5}{2 \cdot 5}

v = \frac{- ( - 10 ) + 4 5}{2 \cdot 5} : \frac{5 + 2 5}{5}

\frac{- ( - 10 ) + 4 5}{2 \cdot 5}

Apply rule

- (- a) = a

= \frac{10 + 4 5}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{10 + 4 5}{10}

Factor

10 + 45 : 2 (5 + 25)

10 + 45

Rewrite as

= 2 \cdot 5 + 2 \cdot 25

Factor out common term

2

= 2 (5 + 25)

= \frac{2 ( 5 + 2 5 )}{10}

Cancel the common factor:

2

= \frac{5 + 2 5}{5}

v = \frac{- ( - 10 ) - 4 5}{2 \cdot 5} : \frac{5 - 2 5}{5}

\frac{- ( - 10 ) - 4 5}{2 \cdot 5}

Apply rule

- (- a) = a

= \frac{10 - 4 5}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{10 - 4 5}{10}

Factor

10 - 45 : 2 (5 - 25)

10 - 45

Rewrite as

= 2 \cdot 5 - 2 \cdot 25

Factor out common term

2

= 2 (5 - 25)

= \frac{2 ( 5 - 2 5 )}{10}

Cancel the common factor:

2

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

The solutions to the quadratic equation are:

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

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

Substitute back

v = u^{2},

solve for

u

Solve

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

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

For

x^{2} = f (a)

the solutions are

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

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

Solve

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

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

For

x^{2} = f (a)

the solutions are

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

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

The solutions are

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

Substitute back

u = tan (x)

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

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

tan (x) = \frac{5 + 2 5}{5} : x = arctan \frac{5 + 2 5}{5} + πn

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

Apply trig inverse properties

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

General solutions for

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

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

x = arctan \frac{5 + 2 5}{5} + πn

x = arctan \frac{5 + 2 5}{5} + πn

tan (x) = - \frac{5 + 2 5}{5} : x = arctan - \frac{5 + 2 5}{5} + πn

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

Apply trig inverse properties

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

General solutions for

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

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

x = arctan - \frac{5 + 2 5}{5} + πn

x = arctan - \frac{5 + 2 5}{5} + πn

tan (x) = \frac{5 - 2 5}{5} : x = arctan \frac{5 - 2 5}{5} + πn

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

Apply trig inverse properties

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

General solutions for

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

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

x = arctan \frac{5 - 2 5}{5} + πn

x = arctan \frac{5 - 2 5}{5} + πn

tan (x) = - \frac{5 - 2 5}{5} : x = arctan - \frac{5 - 2 5}{5} + πn

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

Apply trig inverse properties

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

General solutions for

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

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

x = arctan - \frac{5 - 2 5}{5} + πn

x = arctan - \frac{5 - 2 5}{5} + πn

Combine all the solutions

x = arctan \frac{5 + 2 5}{5} + πn, x = arctan - \frac{5 + 2 5}{5} + πn, x = arctan \frac{5 - 2 5}{5} + πn, x = arctan - \frac{5 - 2 5}{5} + πn

Show solutions in decimal form

x = 0.94247 \dots + πn, x = - 0.94247 \dots + πn, x = 0.31415 \dots + πn, x = - 0.31415 \dots + πn

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