What is the general solution for cos^2(x)-9cos(x)-1=0 ?

The general solution for cos^2(x)-9cos(x)-1=0 is x=1.68079…+2pin,x=-1.68079…+2pin

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cos^2(x)-9cos(x)-1=0

Solution

cos^{2} (x) - 9 cos (x) - 1 = 0

Solution

x = 1.68079 \dots + 2 πn, x = - 1.68079 \dots + 2 πn

Degrees

x = 96.30218 \dots^{\circ} + 36 0^{\circ} n, x = - 96.30218 \dots^{\circ} + 36 0^{\circ} n

Solution steps

cos^{2} (x) - 9 cos (x) - 1 = 0

Solve by substitution

cos^{2} (x) - 9 cos (x) - 1 = 0

Let:

cos (x) = u

u^{2} - 9 u - 1 = 0

u^{2} - 9 u - 1 = 0 : u = \frac{9 + 85}{2}, u = \frac{9 - 85}{2}

u^{2} - 9 u - 1 = 0

Solve with the quadratic formula

u^{2} - 9 u - 1 = 0

Quadratic Equation Formula:

For

a = 1, b = - 9, c = - 1

u_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

(- 9)^{2} - 4 \cdot 1 \cdot (- 1) = 85

(- 9)^{2} - 4 \cdot 1 \cdot (- 1)

Apply rule

- (- a) = a

= (- 9)^{2} + 4 \cdot 1 \cdot 1

Apply exponent rule:

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

n

is even

(- 9)^{2} = 9^{2}

= 9^{2} + 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 9^{2} + 4

9^{2} = 81

= 81 + 4

Add the numbers:

81 + 4 = 85

= 85

u_{1, 2} = \frac{- ( - 9 ) \pm 85}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 9 ) + 85}{2 \cdot 1}, u_{2} = \frac{- ( - 9 ) - 85}{2 \cdot 1}

u = \frac{- ( - 9 ) + 85}{2 \cdot 1} : \frac{9 + 85}{2}

\frac{- ( - 9 ) + 85}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{9 + 85}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{9 + 85}{2}

u = \frac{- ( - 9 ) - 85}{2 \cdot 1} : \frac{9 - 85}{2}

\frac{- ( - 9 ) - 85}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{9 - 85}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{9 - 85}{2}

The solutions to the quadratic equation are:

u = \frac{9 + 85}{2}, u = \frac{9 - 85}{2}

Substitute back

u = cos (x)

cos (x) = \frac{9 + 85}{2}, cos (x) = \frac{9 - 85}{2}

cos (x) = \frac{9 + 85}{2}, cos (x) = \frac{9 - 85}{2}

cos (x) = \frac{9 + 85}{2} :

No Solution

cos (x) = \frac{9 + 85}{2}

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = \frac{9 - 85}{2} : x = arccos (\frac{9 - 85}{2}) + 2 πn, x = - arccos (\frac{9 - 85}{2}) + 2 πn

cos (x) = \frac{9 - 85}{2}

Apply trig inverse properties

cos (x) = \frac{9 - 85}{2}

General solutions for

cos (x) = \frac{9 - 85}{2}

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

x = arccos (\frac{9 - 85}{2}) + 2 πn, x = - arccos (\frac{9 - 85}{2}) + 2 πn

x = arccos (\frac{9 - 85}{2}) + 2 πn, x = - arccos (\frac{9 - 85}{2}) + 2 πn

Combine all the solutions

x = arccos (\frac{9 - 85}{2}) + 2 πn, x = - arccos (\frac{9 - 85}{2}) + 2 πn

Show solutions in decimal form

x = 1.68079 \dots + 2 πn, x = - 1.68079 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for cos^2(x)-9cos(x)-1=0 ?
The general solution for cos^2(x)-9cos(x)-1=0 is x=1.68079…+2pin,x=-1.68079…+2pin