What is the general solution for 3cos^2(x)-8cos(x)-3=0 ?

The general solution for 3cos^2(x)-8cos(x)-3=0 is x=1.91063…+2pin,x=-1.91063…+2pin

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3cos^2(x)-8cos(x)-3=0

Solution

3 cos^{2} (x) - 8 cos (x) - 3 = 0

Solution

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

Degrees

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

Solution steps

3 cos^{2} (x) - 8 cos (x) - 3 = 0

Solve by substitution

3 cos^{2} (x) - 8 cos (x) - 3 = 0

Let:

cos (x) = u

3 u^{2} - 8 u - 3 = 0

3 u^{2} - 8 u - 3 = 0 : u = 3, u = - \frac{1}{3}

3 u^{2} - 8 u - 3 = 0

Solve with the quadratic formula

3 u^{2} - 8 u - 3 = 0

Quadratic Equation Formula:

For

a = 3, b = - 8, c = - 3

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 3 ( - 3 )}{2 \cdot 3}

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 3 ( - 3 )}{2 \cdot 3}

(- 8)^{2} - 4 \cdot 3 (- 3) = 10

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

(- 8)^{2} = 8^{2}

= 8^{2} + 4 \cdot 3 \cdot 3

Multiply the numbers:

4 \cdot 3 \cdot 3 = 36

= 8^{2} + 36

8^{2} = 64

= 64 + 36

Add the numbers:

64 + 36 = 100

= 100

Factor the number:

100 = 1 0^{2}

= 1 0^{2}

Apply radical rule:

1 0^{2} = 10

= 10

u_{1, 2} = \frac{- ( - 8 ) \pm 10}{2 \cdot 3}

Separate the solutions

u_{1} = \frac{- ( - 8 ) + 10}{2 \cdot 3}, u_{2} = \frac{- ( - 8 ) - 10}{2 \cdot 3}

u = \frac{- ( - 8 ) + 10}{2 \cdot 3} : 3

\frac{- ( - 8 ) + 10}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{8 + 10}{2 \cdot 3}

Add the numbers:

8 + 10 = 18

= \frac{18}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{18}{6}

Divide the numbers:

\frac{18}{6} = 3

= 3

u = \frac{- ( - 8 ) - 10}{2 \cdot 3} : - \frac{1}{3}

\frac{- ( - 8 ) - 10}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{8 - 10}{2 \cdot 3}

Subtract the numbers:

8 - 10 = - 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 = 3, u = - \frac{1}{3}

Substitute back

u = cos (x)

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

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

cos (x) = 3 :

No Solution

cos (x) = 3

- 1 \leq cos (x) \leq 1

No Solution

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

Show solutions in decimal form

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

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Frequently Asked Questions (FAQ)

What is the general solution for 3cos^2(x)-8cos(x)-3=0 ?
The general solution for 3cos^2(x)-8cos(x)-3=0 is x=1.91063…+2pin,x=-1.91063…+2pin