What is the general solution for cos^2(x)-4cos(x)-5=0 ?

The general solution for cos^2(x)-4cos(x)-5=0 is x=pi+2pin

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cos^2(x)-4cos(x)-5=0

Solution

cos^{2} (x) - 4 cos (x) - 5 = 0

Solution

x = π + 2 πn

Degrees

x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

cos^{2} (x) - 4 cos (x) - 5 = 0

Solve by substitution

cos^{2} (x) - 4 cos (x) - 5 = 0

Let:

cos (x) = u

u^{2} - 4 u - 5 = 0

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

u^{2} - 4 u - 5 = 0

Solve with the quadratic formula

u^{2} - 4 u - 5 = 0

Quadratic Equation Formula:

For

a = 1, b = - 4, c = - 5

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

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

(- 4)^{2} - 4 \cdot 1 \cdot (- 5) = 6

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 1 \cdot 5 = 20

= 4^{2} + 20

4^{2} = 16

= 16 + 20

Add the numbers:

16 + 20 = 36

= 36

Factor the number:

36 = 6^{2}

= 6^{2}

Apply radical rule:

6^{2} = 6

= 6

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

Separate the solutions

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

u = \frac{- ( - 4 ) + 6}{2 \cdot 1} : 5

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

Apply rule

- (- a) = a

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

Add the numbers:

4 + 6 = 10

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{10}{2}

Divide the numbers:

\frac{10}{2} = 5

= 5

u = \frac{- ( - 4 ) - 6}{2 \cdot 1} : - 1

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

Apply rule

- (- a) = a

= \frac{4 - 6}{2 \cdot 1}

Subtract the numbers:

4 - 6 = - 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = 5, u = - 1

Substitute back

u = cos (x)

cos (x) = 5, cos (x) = - 1

cos (x) = 5, cos (x) = - 1

cos (x) = 5 :

No Solution

cos (x) = 5

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

General solutions for

cos (x) = - 1

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = π + 2 πn

x = π + 2 πn

Combine all the solutions

x = π + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for cos^2(x)-4cos(x)-5=0 ?
The general solution for cos^2(x)-4cos(x)-5=0 is x=pi+2pin