What is the general solution for 5cos^2(x)+3cos(x)-2=0 ?

The general solution for 5cos^2(x)+3cos(x)-2=0 is x=1.15927…+2pin,x=2pi-1.15927…+2pin,x=pi+2pin

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5cos^2(x)+3cos(x)-2=0

Solution

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

Solution

x = 1.15927 \dots + 2 πn, x = 2 π - 1.15927 \dots + 2 πn, x = π + 2 πn

Degrees

x = 66.42182 \dots^{\circ} + 36 0^{\circ} n, x = 293.57817 \dots^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

5 u^{2} + 3 u - 2 = 0

5 u^{2} + 3 u - 2 = 0 : u = \frac{2}{5}, u = - 1

5 u^{2} + 3 u - 2 = 0

Solve with the quadratic formula

5 u^{2} + 3 u - 2 = 0

Quadratic Equation Formula:

For

a = 5, b = 3, c = - 2

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 5 ( - 2 )}{2 \cdot 5}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 5 ( - 2 )}{2 \cdot 5}

3^{2} - 4 \cdot 5 (- 2) = 7

3^{2} - 4 \cdot 5 (- 2)

Apply rule

- (- a) = a

= 3^{2} + 4 \cdot 5 \cdot 2

Multiply the numbers:

4 \cdot 5 \cdot 2 = 40

= 3^{2} + 40

3^{2} = 9

= 9 + 40

Add the numbers:

9 + 40 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- 3 \pm 7}{2 \cdot 5}

Separate the solutions

u_{1} = \frac{- 3 + 7}{2 \cdot 5}, u_{2} = \frac{- 3 - 7}{2 \cdot 5}

u = \frac{- 3 + 7}{2 \cdot 5} : \frac{2}{5}

\frac{- 3 + 7}{2 \cdot 5}

Add/Subtract the numbers:

- 3 + 7 = 4

= \frac{4}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{4}{10}

Cancel the common factor:

2

= \frac{2}{5}

u = \frac{- 3 - 7}{2 \cdot 5} : - 1

\frac{- 3 - 7}{2 \cdot 5}

Subtract the numbers:

- 3 - 7 = - 10

= \frac{- 10}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{- 10}{10}

Apply the fraction rule:

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

= - \frac{10}{10}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = \frac{2}{5}, u = - 1

Substitute back

u = cos (x)

cos (x) = \frac{2}{5}, cos (x) = - 1

cos (x) = \frac{2}{5}, cos (x) = - 1

cos (x) = \frac{2}{5} : x = arccos (\frac{2}{5}) + 2 πn, x = 2 π - arccos (\frac{2}{5}) + 2 πn

cos (x) = \frac{2}{5}

Apply trig inverse properties

cos (x) = \frac{2}{5}

General solutions for

cos (x) = \frac{2}{5}

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

x = arccos (\frac{2}{5}) + 2 πn, x = 2 π - arccos (\frac{2}{5}) + 2 πn

x = arccos (\frac{2}{5}) + 2 πn, x = 2 π - arccos (\frac{2}{5}) + 2 πn

cos (x) = - 1 : x = π + 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 = π + 2 πn

x = π + 2 πn

Combine all the solutions

x = arccos (\frac{2}{5}) + 2 πn, x = 2 π - arccos (\frac{2}{5}) + 2 πn, x = π + 2 πn

Show solutions in decimal form

x = 1.15927 \dots + 2 πn, x = 2 π - 1.15927 \dots + 2 πn, x = π + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 5cos^2(x)+3cos(x)-2=0 ?
The general solution for 5cos^2(x)+3cos(x)-2=0 is x=1.15927…+2pin,x=2pi-1.15927…+2pin,x=pi+2pin