What is the general solution for 8cos^2(x)+14cos(x)=-3 ?

The general solution for 8cos^2(x)+14cos(x)=-3 is x=1.82347…+2pin,x=-1.82347…+2pin

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8cos^2(x)+14cos(x)=-3

Solution

8 cos^{2} (x) + 14 cos (x) = - 3

Solution

x = 1.82347 \dots + 2 πn, x = - 1.82347 \dots + 2 πn

Degrees

x = 104.47751 \dots^{\circ} + 36 0^{\circ} n, x = - 104.47751 \dots^{\circ} + 36 0^{\circ} n

Solution steps

8 cos^{2} (x) + 14 cos (x) = - 3

Solve by substitution

8 cos^{2} (x) + 14 cos (x) = - 3

Let:

cos (x) = u

8 u^{2} + 14 u = - 3

8 u^{2} + 14 u = - 3 : u = - \frac{1}{4}, u = - \frac{3}{2}

8 u^{2} + 14 u = - 3

Move

3

to the left side

8 u^{2} + 14 u = - 3

Add

3

to both sides

8 u^{2} + 14 u + 3 = - 3 + 3

Simplify

8 u^{2} + 14 u + 3 = 0

8 u^{2} + 14 u + 3 = 0

Solve with the quadratic formula

8 u^{2} + 14 u + 3 = 0

Quadratic Equation Formula:

For

a = 8, b = 14, c = 3

u_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 8 \cdot 3}{2 \cdot 8}

u_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 8 \cdot 3}{2 \cdot 8}

1 4^{2} - 4 \cdot 8 \cdot 3 = 10

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

Multiply the numbers:

4 \cdot 8 \cdot 3 = 96

= 1 4^{2} - 96

1 4^{2} = 196

= 196 - 96

Subtract the numbers:

196 - 96 = 100

= 100

Factor the number:

100 = 1 0^{2}

= 1 0^{2}

Apply radical rule:

1 0^{2} = 10

= 10

u_{1, 2} = \frac{- 14 \pm 10}{2 \cdot 8}

Separate the solutions

u_{1} = \frac{- 14 + 10}{2 \cdot 8}, u_{2} = \frac{- 14 - 10}{2 \cdot 8}

u = \frac{- 14 + 10}{2 \cdot 8} : - \frac{1}{4}

\frac{- 14 + 10}{2 \cdot 8}

Add/Subtract the numbers:

- 14 + 10 = - 4

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

Multiply the numbers:

2 \cdot 8 = 16

= \frac{- 4}{16}

Apply the fraction rule:

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

= - \frac{4}{16}

Cancel the common factor:

4

= - \frac{1}{4}

u = \frac{- 14 - 10}{2 \cdot 8} : - \frac{3}{2}

\frac{- 14 - 10}{2 \cdot 8}

Subtract the numbers:

- 14 - 10 = - 24

= \frac{- 24}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{- 24}{16}

Apply the fraction rule:

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

= - \frac{24}{16}

Cancel the common factor:

8

= - \frac{3}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

cos (x) = - \frac{1}{4}, cos (x) = - \frac{3}{2}

cos (x) = - \frac{1}{4}, cos (x) = - \frac{3}{2}

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

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

Apply trig inverse properties

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

General solutions for

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

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

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

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

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

No Solution

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

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

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

Show solutions in decimal form

x = 1.82347 \dots + 2 πn, x = - 1.82347 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 8cos^2(x)+14cos(x)=-3 ?
The general solution for 8cos^2(x)+14cos(x)=-3 is x=1.82347…+2pin,x=-1.82347…+2pin