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

The general solution for cos^2(x)-4cos(x)-1=0 is x=1.80911…+2pin,x=-1.80911…+2pin

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

Solution

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

Solution

x = 1.80911 \dots + 2 πn, x = - 1.80911 \dots + 2 πn

Degrees

x = 103.65458 \dots^{\circ} + 36 0^{\circ} n, x = - 103.65458 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

(- 4)^{2} - 4 \cdot 1 \cdot (- 1) = 25

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 4^{2} + 4

4^{2} = 16

= 16 + 4

Add the numbers:

16 + 4 = 20

= 20

Prime factorization of

20 : 2^{2} \cdot 5

20

20

divides by

220 = 10 \cdot 2

= 2 \cdot 10

10

divides by

210 = 5 \cdot 2

= 2 \cdot 2 \cdot 5

2, 5

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 5

= 2^{2} \cdot 5

= 2^{2} \cdot 5

Apply radical rule:

= 5 2^{2}

Apply radical rule:

2^{2} = 2

= 25

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

Separate the solutions

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

u = \frac{- ( - 4 ) + 2 5}{2 \cdot 1} : 2 + 5

\frac{- ( - 4 ) + 2 5}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{4 + 2 5}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

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

Factor

4 + 25 : 2 (2 + 5)

4 + 25

Rewrite as

= 2 \cdot 2 + 25

Factor out common term

2

= 2 (2 + 5)

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

Divide the numbers:

\frac{2}{2} = 1

= 2 + 5

u = \frac{- ( - 4 ) - 2 5}{2 \cdot 1} : 2 - 5

\frac{- ( - 4 ) - 2 5}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{4 - 2 5}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

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

Factor

4 - 25 : 2 (2 - 5)

4 - 25

Rewrite as

= 2 \cdot 2 - 25

Factor out common term

2

= 2 (2 - 5)

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

Divide the numbers:

\frac{2}{2} = 1

= 2 - 5

The solutions to the quadratic equation are:

u = 2 + 5, u = 2 - 5

Substitute back

u = cos (x)

cos (x) = 2 + 5, cos (x) = 2 - 5

cos (x) = 2 + 5, cos (x) = 2 - 5

cos (x) = 2 + 5 :

No Solution

cos (x) = 2 + 5

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = 2 - 5 : x = arccos (2 - 5) + 2 πn, x = - arccos (2 - 5) + 2 πn

cos (x) = 2 - 5

Apply trig inverse properties

cos (x) = 2 - 5

General solutions for

cos (x) = 2 - 5

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos (2 - 5) + 2 πn, x = - arccos (2 - 5) + 2 πn

x = arccos (2 - 5) + 2 πn, x = - arccos (2 - 5) + 2 πn

Combine all the solutions

x = arccos (2 - 5) + 2 πn, x = - arccos (2 - 5) + 2 πn

Show solutions in decimal form

x = 1.80911 \dots + 2 πn, x = - 1.80911 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for cos^2(x)-4cos(x)-1=0 ?
The general solution for cos^2(x)-4cos(x)-1=0 is x=1.80911…+2pin,x=-1.80911…+2pin