What is the general solution for cos^2(x)-6cos(x)-1=0 ?

The general solution for cos^2(x)-6cos(x)-1=0 is x=1.73379…+2pin,x=-1.73379…+2pin

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cos^2(x)-6cos(x)-1=0

Solution

cos^{2} (x) - 6 cos (x) - 1 = 0

Solution

x = 1.73379 \dots + 2 πn, x = - 1.73379 \dots + 2 πn

Degrees

x = 99.33912 \dots^{\circ} + 36 0^{\circ} n, x = - 99.33912 \dots^{\circ} + 36 0^{\circ} n

Solution steps

cos^{2} (x) - 6 cos (x) - 1 = 0

Solve by substitution

cos^{2} (x) - 6 cos (x) - 1 = 0

Let:

cos (x) = u

u^{2} - 6 u - 1 = 0

u^{2} - 6 u - 1 = 0 : u = 3 + 10, u = 3 - 10

u^{2} - 6 u - 1 = 0

Solve with the quadratic formula

u^{2} - 6 u - 1 = 0

Quadratic Equation Formula:

For

a = 1, b = - 6, c = - 1

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

(- 6)^{2} - 4 \cdot 1 \cdot (- 1) = 210

(- 6)^{2} - 4 \cdot 1 \cdot (- 1)

Apply rule

- (- a) = a

= (- 6)^{2} + 4 \cdot 1 \cdot 1

Apply exponent rule:

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

n

is even

(- 6)^{2} = 6^{2}

= 6^{2} + 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 6^{2} + 4

6^{2} = 36

= 36 + 4

Add the numbers:

36 + 4 = 40

= 40

Prime factorization of

40 : 2^{3} \cdot 5

40

40

divides by

240 = 20 \cdot 2

= 2 \cdot 20

20

divides by

220 = 10 \cdot 2

= 2 \cdot 2 \cdot 10

10

divides by

210 = 5 \cdot 2

= 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 5

= 2^{3} \cdot 5

= 2^{3} \cdot 5

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{2} \cdot 2 \cdot 5

Apply radical rule:

= 2^{2} 2 \cdot 5

Apply radical rule:

2^{2} = 2

= 2 2 \cdot 5

Refine

= 210

u_{1, 2} = \frac{- ( - 6 ) \pm 2 10}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 6 ) + 2 10}{2 \cdot 1}, u_{2} = \frac{- ( - 6 ) - 2 10}{2 \cdot 1}

u = \frac{- ( - 6 ) + 2 10}{2 \cdot 1} : 3 + 10

\frac{- ( - 6 ) + 2 10}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{6 + 2 10}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{6 + 2 10}{2}

Factor

6 + 210 : 2 (3 + 10)

6 + 210

Rewrite as

= 2 \cdot 3 + 210

Factor out common term

2

= 2 (3 + 10)

= \frac{2 ( 3 + 10 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= 3 + 10

u = \frac{- ( - 6 ) - 2 10}{2 \cdot 1} : 3 - 10

\frac{- ( - 6 ) - 2 10}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{6 - 2 10}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{6 - 2 10}{2}

Factor

6 - 210 : 2 (3 - 10)

6 - 210

Rewrite as

= 2 \cdot 3 - 210

Factor out common term

2

= 2 (3 - 10)

= \frac{2 ( 3 - 10 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= 3 - 10

The solutions to the quadratic equation are:

u = 3 + 10, u = 3 - 10

Substitute back

u = cos (x)

cos (x) = 3 + 10, cos (x) = 3 - 10

cos (x) = 3 + 10, cos (x) = 3 - 10

cos (x) = 3 + 10 :

No Solution

cos (x) = 3 + 10

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = 3 - 10 : x = arccos (3 - 10) + 2 πn, x = - arccos (3 - 10) + 2 πn

cos (x) = 3 - 10

Apply trig inverse properties

cos (x) = 3 - 10

General solutions for

cos (x) = 3 - 10

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

x = arccos (3 - 10) + 2 πn, x = - arccos (3 - 10) + 2 πn

x = arccos (3 - 10) + 2 πn, x = - arccos (3 - 10) + 2 πn

Combine all the solutions

x = arccos (3 - 10) + 2 πn, x = - arccos (3 - 10) + 2 πn

Show solutions in decimal form

x = 1.73379 \dots + 2 πn, x = - 1.73379 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for cos^2(x)-6cos(x)-1=0 ?
The general solution for cos^2(x)-6cos(x)-1=0 is x=1.73379…+2pin,x=-1.73379…+2pin