What is the general solution for 2cos^2(x)+9cos(x)+4=0 ?

The general solution for 2cos^2(x)+9cos(x)+4=0 is x=(2pi)/3+2pin,x=(4pi)/3+2pin

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2cos^2(x)+9cos(x)+4=0

Solution

2 cos^{2} (x) + 9 cos (x) + 4 = 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

2 cos^{2} (x) + 9 cos (x) + 4 = 0

Solve by substitution

2 cos^{2} (x) + 9 cos (x) + 4 = 0

Let:

cos (x) = u

2 u^{2} + 9 u + 4 = 0

2 u^{2} + 9 u + 4 = 0 : u = - \frac{1}{2}, u = - 4

2 u^{2} + 9 u + 4 = 0

Solve with the quadratic formula

2 u^{2} + 9 u + 4 = 0

Quadratic Equation Formula:

For

a = 2, b = 9, c = 4

u_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 2 \cdot 4}{2 \cdot 2}

u_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 2 \cdot 4}{2 \cdot 2}

9^{2} - 4 \cdot 2 \cdot 4 = 7

9^{2} - 4 \cdot 2 \cdot 4

Multiply the numbers:

4 \cdot 2 \cdot 4 = 32

= 9^{2} - 32

9^{2} = 81

= 81 - 32

Subtract the numbers:

81 - 32 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- 9 \pm 7}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- 9 + 7}{2 \cdot 2}, u_{2} = \frac{- 9 - 7}{2 \cdot 2}

u = \frac{- 9 + 7}{2 \cdot 2} : - \frac{1}{2}

\frac{- 9 + 7}{2 \cdot 2}

Add/Subtract the numbers:

- 9 + 7 = - 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2}{4}

Apply the fraction rule:

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

= - \frac{2}{4}

Cancel the common factor:

2

= - \frac{1}{2}

u = \frac{- 9 - 7}{2 \cdot 2} : - 4

\frac{- 9 - 7}{2 \cdot 2}

Subtract the numbers:

- 9 - 7 = - 16

= \frac{- 16}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 16}{4}

Apply the fraction rule:

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

= - \frac{16}{4}

Divide the numbers:

\frac{16}{4} = 4

= - 4

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

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

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

cos (x) = - 4 :

No Solution

cos (x) = - 4

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

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

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(x)+9cos(x)+4=0 ?
The general solution for 2cos^2(x)+9cos(x)+4=0 is x=(2pi)/3+2pin,x=(4pi)/3+2pin