What is the general solution for 4cos^2(x)-4cos(x)-1=0 ?

The general solution for 4cos^2(x)-4cos(x)-1=0 is x=1.77941…+2pin,x=-1.77941…+2pin

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

Solution

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

Solution

x = 1.77941 \dots + 2 πn, x = - 1.77941 \dots + 2 πn

Degrees

x = 101.95285 \dots^{\circ} + 36 0^{\circ} n, x = - 101.95285 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

4 u^{2} - 4 u - 1 = 0

4 u^{2} - 4 u - 1 = 0 : u = \frac{1 + 2}{2}, u = \frac{1 - 2}{2}

4 u^{2} - 4 u - 1 = 0

Solve with the quadratic formula

4 u^{2} - 4 u - 1 = 0

Quadratic Equation Formula:

For

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

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

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

(- 4)^{2} - 4 \cdot 4 (- 1) = 42

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 4^{2} + 16

4^{2} = 16

= 16 + 16

Add the numbers:

16 + 16 = 32

= 32

Prime factorization of

32 : 2^{5}

32

32

divides by

232 = 16 \cdot 2

= 2 \cdot 16

16

divides by

216 = 8 \cdot 2

= 2 \cdot 2 \cdot 8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

= 2^{5}

= 2^{5}

Apply exponent rule:

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

= 2^{4} \cdot 2

Apply radical rule:

= 2 2^{4}

Apply radical rule:

2^{4} = 2^{\frac{4}{2}} = 2^{2}

= 2^{2} 2

Refine

= 42

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{4 + 4 2}{8}

Factor

4 + 42 : 4 (1 + 2)

4 + 42

Rewrite as

= 4 \cdot 1 + 42

Factor out common term

4

= 4 (1 + 2)

= \frac{4 ( 1 + 2 )}{8}

Cancel the common factor:

4

= \frac{1 + 2}{2}

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{4 - 4 2}{8}

Factor

4 - 42 : 4 (1 - 2)

4 - 42

Rewrite as

= 4 \cdot 1 - 42

Factor out common term

4

= 4 (1 - 2)

= \frac{4 ( 1 - 2 )}{8}

Cancel the common factor:

4

= \frac{1 - 2}{2}

The solutions to the quadratic equation are:

u = \frac{1 + 2}{2}, u = \frac{1 - 2}{2}

Substitute back

u = cos (x)

cos (x) = \frac{1 + 2}{2}, cos (x) = \frac{1 - 2}{2}

cos (x) = \frac{1 + 2}{2}, cos (x) = \frac{1 - 2}{2}

cos (x) = \frac{1 + 2}{2} :

No Solution

cos (x) = \frac{1 + 2}{2}

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = \frac{1 - 2}{2} : x = arccos (\frac{1 - 2}{2}) + 2 πn, x = - arccos (\frac{1 - 2}{2}) + 2 πn

cos (x) = \frac{1 - 2}{2}

Apply trig inverse properties

cos (x) = \frac{1 - 2}{2}

General solutions for

cos (x) = \frac{1 - 2}{2}

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

x = arccos (\frac{1 - 2}{2}) + 2 πn, x = - arccos (\frac{1 - 2}{2}) + 2 πn

x = arccos (\frac{1 - 2}{2}) + 2 πn, x = - arccos (\frac{1 - 2}{2}) + 2 πn

Combine all the solutions

x = arccos (\frac{1 - 2}{2}) + 2 πn, x = - arccos (\frac{1 - 2}{2}) + 2 πn

Show solutions in decimal form

x = 1.77941 \dots + 2 πn, x = - 1.77941 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4cos^2(x)-4cos(x)-1=0 ?
The general solution for 4cos^2(x)-4cos(x)-1=0 is x=1.77941…+2pin,x=-1.77941…+2pin