What is the general solution for cos^2(x)-8cos(x)-1=0 ?

The general solution for cos^2(x)-8cos(x)-1=0 is x=1.69421…+2pin,x=-1.69421…+2pin

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cos^2(x)-8cos(x)-1=0

Solution

cos^{2} (x) - 8 cos (x) - 1 = 0

Solution

x = 1.69421 \dots + 2 πn, x = - 1.69421 \dots + 2 πn

Degrees

x = 97.07137 \dots^{\circ} + 36 0^{\circ} n, x = - 97.07137 \dots^{\circ} + 36 0^{\circ} n

Solution steps

cos^{2} (x) - 8 cos (x) - 1 = 0

Solve by substitution

cos^{2} (x) - 8 cos (x) - 1 = 0

Let:

cos (x) = u

u^{2} - 8 u - 1 = 0

u^{2} - 8 u - 1 = 0 : u = 4 + 17, u = 4 - 17

u^{2} - 8 u - 1 = 0

Solve with the quadratic formula

u^{2} - 8 u - 1 = 0

Quadratic Equation Formula:

For

a = 1, b = - 8, c = - 1

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

(- 8)^{2} - 4 \cdot 1 \cdot (- 1) = 217

(- 8)^{2} - 4 \cdot 1 \cdot (- 1)

Apply rule

- (- a) = a

= (- 8)^{2} + 4 \cdot 1 \cdot 1

Apply exponent rule:

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

n

is even

(- 8)^{2} = 8^{2}

= 8^{2} + 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 8^{2} + 4

8^{2} = 64

= 64 + 4

Add the numbers:

64 + 4 = 68

= 68

Prime factorization of

68 : 2^{2} \cdot 17

68

68

divides by

268 = 34 \cdot 2

= 2 \cdot 34

34

divides by

234 = 17 \cdot 2

= 2 \cdot 2 \cdot 17

2, 17

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 17

= 2^{2} \cdot 17

= 2^{2} \cdot 17

Apply radical rule:

= 17 2^{2}

Apply radical rule:

2^{2} = 2

= 217

u_{1, 2} = \frac{- ( - 8 ) \pm 2 17}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 8 ) + 2 17}{2 \cdot 1}, u_{2} = \frac{- ( - 8 ) - 2 17}{2 \cdot 1}

u = \frac{- ( - 8 ) + 2 17}{2 \cdot 1} : 4 + 17

\frac{- ( - 8 ) + 2 17}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{8 + 2 17}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{8 + 2 17}{2}

Factor

8 + 217 : 2 (4 + 17)

8 + 217

Rewrite as

= 2 \cdot 4 + 217

Factor out common term

2

= 2 (4 + 17)

= \frac{2 ( 4 + 17 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= 4 + 17

u = \frac{- ( - 8 ) - 2 17}{2 \cdot 1} : 4 - 17

\frac{- ( - 8 ) - 2 17}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{8 - 2 17}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{8 - 2 17}{2}

Factor

8 - 217 : 2 (4 - 17)

8 - 217

Rewrite as

= 2 \cdot 4 - 217

Factor out common term

2

= 2 (4 - 17)

= \frac{2 ( 4 - 17 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= 4 - 17

The solutions to the quadratic equation are:

u = 4 + 17, u = 4 - 17

Substitute back

u = cos (x)

cos (x) = 4 + 17, cos (x) = 4 - 17

cos (x) = 4 + 17, cos (x) = 4 - 17

cos (x) = 4 + 17 :

No Solution

cos (x) = 4 + 17

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = 4 - 17 : x = arccos (4 - 17) + 2 πn, x = - arccos (4 - 17) + 2 πn

cos (x) = 4 - 17

Apply trig inverse properties

cos (x) = 4 - 17

General solutions for

cos (x) = 4 - 17

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

x = arccos (4 - 17) + 2 πn, x = - arccos (4 - 17) + 2 πn

x = arccos (4 - 17) + 2 πn, x = - arccos (4 - 17) + 2 πn

Combine all the solutions

x = arccos (4 - 17) + 2 πn, x = - arccos (4 - 17) + 2 πn

Show solutions in decimal form

x = 1.69421 \dots + 2 πn, x = - 1.69421 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for cos^2(x)-8cos(x)-1=0 ?
The general solution for cos^2(x)-8cos(x)-1=0 is x=1.69421…+2pin,x=-1.69421…+2pin