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

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

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

Solution

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

Solution

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Degrees

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

(- 4)^{2} - 4 \cdot 4 (- 3) = 8

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 4 \cdot 3 = 48

= 4^{2} + 48

4^{2} = 16

= 16 + 48

Add the numbers:

16 + 48 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

8^{2} = 8

= 8

u_{1, 2} = \frac{- ( - 4 ) \pm 8}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- ( - 4 ) + 8}{2 \cdot 4}, u_{2} = \frac{- ( - 4 ) - 8}{2 \cdot 4}

u = \frac{- ( - 4 ) + 8}{2 \cdot 4} : \frac{3}{2}

\frac{- ( - 4 ) + 8}{2 \cdot 4}

Apply rule

- (- a) = a

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

Add the numbers:

4 + 8 = 12

= \frac{12}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{12}{8}

Cancel the common factor:

4

= \frac{3}{2}

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

\frac{- ( - 4 ) - 8}{2 \cdot 4}

Apply rule

- (- a) = a

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

Subtract the numbers:

4 - 8 = - 4

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 4}{8}

Apply the fraction rule:

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

= - \frac{4}{8}

Cancel the common factor:

4

= - \frac{1}{2}

The solutions to the quadratic equation are:

u = \frac{3}{2}, u = - \frac{1}{2}

Substitute back

u = cos (x)

cos (x) = \frac{3}{2}, cos (x) = - \frac{1}{2}

cos (x) = \frac{3}{2}, cos (x) = - \frac{1}{2}

cos (x) = \frac{3}{2} :

No Solution

cos (x) = \frac{3}{2}

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

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

General solutions for

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

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 = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Combine all the solutions

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Popular Examples

cot(x)= 4/3

cot (x) = \frac{4}{3}

tan(θ)=-5/2

tan (θ) = - \frac{5}{2}

-1=cos(t)+cos(2t)

- 1 = cos (t) + cos (2 t)

15cos(3x)=0

15 cos (3 x) = 0

-1/2 =sin(2x)

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

Frequently Asked Questions (FAQ)

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