What is the general solution for 3cos^2(x)+10cos(x)+3=0 ?

The general solution for 3cos^2(x)+10cos(x)+3=0 is x=1.91063…+2pin,x=-1.91063…+2pin

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3cos^2(x)+10cos(x)+3=0

Solution

3 cos^{2} (x) + 10 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) + 10 cos (x) + 3 = 0

Solve by substitution

3 cos^{2} (x) + 10 cos (x) + 3 = 0

Let:

cos (x) = u

3 u^{2} + 10 u + 3 = 0

3 u^{2} + 10 u + 3 = 0 : u = - \frac{1}{3}, u = - 3

3 u^{2} + 10 u + 3 = 0

Solve with the quadratic formula

3 u^{2} + 10 u + 3 = 0

Quadratic Equation Formula:

For

a = 3, b = 10, c = 3

u_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 3 \cdot 3}{2 \cdot 3}

u_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 3 \cdot 3}{2 \cdot 3}

1 0^{2} - 4 \cdot 3 \cdot 3 = 8

1 0^{2} - 4 \cdot 3 \cdot 3

Multiply the numbers:

4 \cdot 3 \cdot 3 = 36

= 1 0^{2} - 36

1 0^{2} = 100

= 100 - 36

Subtract the numbers:

100 - 36 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

8^{2} = 8

= 8

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

Separate the solutions

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

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

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

Add/Subtract the numbers:

- 10 + 8 = - 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}

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

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

Subtract the numbers:

- 10 - 8 = - 18

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

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 18}{6}

Apply the fraction rule:

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

= - \frac{18}{6}

Divide the numbers:

\frac{18}{6} = 3

= - 3

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

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

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

cos (x) = - 3 :

No Solution

cos (x) = - 3

- 1 \leq cos (x) \leq 1

No Solution

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)+10cos(x)+3=0 ?
The general solution for 3cos^2(x)+10cos(x)+3=0 is x=1.91063…+2pin,x=-1.91063…+2pin