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=1.84206…+2pin,x=-1.84206…+2pin

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Popular >

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

Solution

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

Solution

x = 1.84206 \dots + 2 πn, x = - 1.84206 \dots + 2 πn

Degrees

x = 105.54226 \dots^{\circ} + 36 0^{\circ} n, x = - 105.54226 \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 + 3, u = - 2 - 3

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 = 23

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

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 4^{2} - 4

4^{2} = 16

= 16 - 4

Subtract the numbers:

16 - 4 = 12

= 12

Prime factorization of

12 : 2^{2} \cdot 3

12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3

= 2^{2} \cdot 3

= 2^{2} \cdot 3

Apply radical rule:

n ab = n a n b

= 3 2^{2}

Apply radical rule:

n a^{n} = a

2^{2} = 2

= 23

u_{1, 2} = \frac{- 4 \pm 2 3}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 4 + 2 3}{2 \cdot 1}, u_{2} = \frac{- 4 - 2 3}{2 \cdot 1}

u = \frac{- 4 + 2 3}{2 \cdot 1} : - 2 + 3

\frac{- 4 + 2 3}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 4 + 2 3}{2}

Factor

- 4 + 23 : 2 (- 2 + 3)

- 4 + 23

Rewrite as

= - 2 \cdot 2 + 23

Factor out common term

2

= 2 (- 2 + 3)

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

Divide the numbers:

\frac{2}{2} = 1

= - 2 + 3

u = \frac{- 4 - 2 3}{2 \cdot 1} : - 2 - 3

\frac{- 4 - 2 3}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 4 - 2 3}{2}

Factor

- 4 - 23 : - 2 (2 + 3)

- 4 - 23

Rewrite as

= - 2 \cdot 2 - 23

Factor out common term

2

= - 2 (2 + 3)

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

Divide the numbers:

\frac{2}{2} = 1

= - (2 + 3)

Negate

- (2 + 3) = - 2 - 3

= - 2 - 3

The solutions to the quadratic equation are:

u = - 2 + 3, u = - 2 - 3

Substitute back

u = cos (x)

cos (x) = - 2 + 3, cos (x) = - 2 - 3

cos (x) = - 2 + 3, cos (x) = - 2 - 3

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

cos (x) = - 2 + 3

Apply trig inverse properties

cos (x) = - 2 + 3

General solutions for

cos (x) = - 2 + 3

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

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

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

cos (x) = - 2 - 3 :

No Solution

cos (x) = - 2 - 3

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

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

Show solutions in decimal form

x = 1.84206 \dots + 2 πn, x = - 1.84206 \dots + 2 πn

Graph

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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=1.84206…+2pin,x=-1.84206…+2pin