What is the general solution for 3cos^2(x)-2cos(x)-1=0 ?

The general solution for 3cos^2(x)-2cos(x)-1=0 is x=2pin,x=1.91063…+2pin,x=-1.91063…+2pin

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3cos^2(x)-2cos(x)-1=0

Solution

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

Solution

x = 2 πn, x = 1.91063 \dots + 2 πn, x = - 1.91063 \dots + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 109.47122 \dots^{\circ} + 36 0^{\circ} n, x = - 109.47122 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 3, b = - 2, c = - 1

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

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

(- 2)^{2} - 4 \cdot 3 (- 1) = 4

(- 2)^{2} - 4 \cdot 3 (- 1)

Apply rule

- (- a) = a

= (- 2)^{2} + 4 \cdot 3 \cdot 1

Apply exponent rule:

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

n

is even

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

= 2^{2} + 4 \cdot 3 \cdot 1

Multiply the numbers:

4 \cdot 3 \cdot 1 = 12

= 2^{2} + 12

2^{2} = 4

= 4 + 12

Add the numbers:

4 + 12 = 16

= 16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

n a^{n} = a

4^{2} = 4

= 4

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

Separate the solutions

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

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

\frac{- ( - 2 ) + 4}{2 \cdot 3}

Apply rule

- (- a) = a

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

Add the numbers:

2 + 4 = 6

= \frac{6}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{6}{6}

Apply rule

\frac{a}{a} = 1

= 1

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

2 - 4 = - 2

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

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 2}{6}

Apply the fraction rule:

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

= - \frac{2}{6}

Cancel the common factor:

2

= - \frac{1}{3}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

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

cos (x) = 1 : x = 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 = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

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

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

Apply trig inverse properties

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

General solutions for

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

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

x = arccos (- \frac{1}{3}) + 2 πn, x = - arccos (- \frac{1}{3}) + 2 πn

x = arccos (- \frac{1}{3}) + 2 πn, x = - arccos (- \frac{1}{3}) + 2 πn

Combine all the solutions

x = 2 πn, x = arccos (- \frac{1}{3}) + 2 πn, x = - arccos (- \frac{1}{3}) + 2 πn

Show solutions in decimal form

x = 2 πn, x = 1.91063 \dots + 2 πn, x = - 1.91063 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 3cos^2(x)-2cos(x)-1=0 ?
The general solution for 3cos^2(x)-2cos(x)-1=0 is x=2pin,x=1.91063…+2pin,x=-1.91063…+2pin