What is the general solution for 4cos^2(x)-2cos(x)-3=0 ?

The general solution for 4cos^2(x)-2cos(x)-3=0 is x=2.28020…+2pin,x=-2.28020…+2pin

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4cos^2(x)-2cos(x)-3=0

Solution

4 cos^{2} (x) - 2 cos (x) - 3 = 0

Solution

x = 2.28020 \dots + 2 πn, x = - 2.28020 \dots + 2 πn

Degrees

x = 130.64631 \dots^{\circ} + 36 0^{\circ} n, x = - 130.64631 \dots^{\circ} + 36 0^{\circ} n

Solution steps

4 cos^{2} (x) - 2 cos (x) - 3 = 0

Solve by substitution

4 cos^{2} (x) - 2 cos (x) - 3 = 0

Let:

cos (x) = u

4 u^{2} - 2 u - 3 = 0

4 u^{2} - 2 u - 3 = 0 : u = \frac{1 + 13}{4}, u = \frac{1 - 13}{4}

4 u^{2} - 2 u - 3 = 0

Solve with the quadratic formula

4 u^{2} - 2 u - 3 = 0

Quadratic Equation Formula:

For

a = 4, b = - 2, c = - 3

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

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

(- 2)^{2} - 4 \cdot 4 (- 3) = 213

(- 2)^{2} - 4 \cdot 4 (- 3)

Apply rule

- (- a) = a

= (- 2)^{2} + 4 \cdot 4 \cdot 3

Apply exponent rule:

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

n

is even

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

= 2^{2} + 4 \cdot 4 \cdot 3

Multiply the numbers:

4 \cdot 4 \cdot 3 = 48

= 2^{2} + 48

2^{2} = 4

= 4 + 48

Add the numbers:

4 + 48 = 52

= 52

Prime factorization of

52 : 2^{2} \cdot 13

52

52

divides by

252 = 26 \cdot 2

= 2 \cdot 26

26

divides by

226 = 13 \cdot 2

= 2 \cdot 2 \cdot 13

2, 13

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 13

= 2^{2} \cdot 13

= 2^{2} \cdot 13

Apply radical rule:

= 13 2^{2}

Apply radical rule:

2^{2} = 2

= 213

u_{1, 2} = \frac{- ( - 2 ) \pm 2 13}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 2 13}{2 \cdot 4}, u_{2} = \frac{- ( - 2 ) - 2 13}{2 \cdot 4}

u = \frac{- ( - 2 ) + 2 13}{2 \cdot 4} : \frac{1 + 13}{4}

\frac{- ( - 2 ) + 2 13}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{2 + 2 13}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{2 + 2 13}{8}

Factor

2 + 213 : 2 (1 + 13)

2 + 213

Rewrite as

= 2 \cdot 1 + 213

Factor out common term

2

= 2 (1 + 13)

= \frac{2 ( 1 + 13 )}{8}

Cancel the common factor:

2

= \frac{1 + 13}{4}

u = \frac{- ( - 2 ) - 2 13}{2 \cdot 4} : \frac{1 - 13}{4}

\frac{- ( - 2 ) - 2 13}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{2 - 2 13}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{2 - 2 13}{8}

Factor

2 - 213 : 2 (1 - 13)

2 - 213

Rewrite as

= 2 \cdot 1 - 213

Factor out common term

2

= 2 (1 - 13)

= \frac{2 ( 1 - 13 )}{8}

Cancel the common factor:

2

= \frac{1 - 13}{4}

The solutions to the quadratic equation are:

u = \frac{1 + 13}{4}, u = \frac{1 - 13}{4}

Substitute back

u = cos (x)

cos (x) = \frac{1 + 13}{4}, cos (x) = \frac{1 - 13}{4}

cos (x) = \frac{1 + 13}{4}, cos (x) = \frac{1 - 13}{4}

cos (x) = \frac{1 + 13}{4} :

No Solution

cos (x) = \frac{1 + 13}{4}

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = \frac{1 - 13}{4} : x = arccos (\frac{1 - 13}{4}) + 2 πn, x = - arccos (\frac{1 - 13}{4}) + 2 πn

cos (x) = \frac{1 - 13}{4}

Apply trig inverse properties

cos (x) = \frac{1 - 13}{4}

General solutions for

cos (x) = \frac{1 - 13}{4}

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

x = arccos (\frac{1 - 13}{4}) + 2 πn, x = - arccos (\frac{1 - 13}{4}) + 2 πn

x = arccos (\frac{1 - 13}{4}) + 2 πn, x = - arccos (\frac{1 - 13}{4}) + 2 πn

Combine all the solutions

x = arccos (\frac{1 - 13}{4}) + 2 πn, x = - arccos (\frac{1 - 13}{4}) + 2 πn

Show solutions in decimal form

x = 2.28020 \dots + 2 πn, x = - 2.28020 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4cos^2(x)-2cos(x)-3=0 ?
The general solution for 4cos^2(x)-2cos(x)-3=0 is x=2.28020…+2pin,x=-2.28020…+2pin