What is the general solution for -5cos^2(x)+4cos(x)+1=0 ?

The general solution for -5cos^2(x)+4cos(x)+1=0 is x=1.77215…+2pin,x=-1.77215…+2pin,x=2pin

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Popular >

-5cos^2(x)+4cos(x)+1=0

Solution

- 5 cos^{2} (x) + 4 cos (x) + 1 = 0

Solution

x = 1.77215 \dots + 2 πn, x = - 1.77215 \dots + 2 πn, x = 2 πn

Degrees

x = 101.53695 \dots^{\circ} + 36 0^{\circ} n, x = - 101.53695 \dots^{\circ} + 36 0^{\circ} n, x = 0^{\circ} + 36 0^{\circ} n

Solution steps

- 5 cos^{2} (x) + 4 cos (x) + 1 = 0

Solve by substitution

- 5 cos^{2} (x) + 4 cos (x) + 1 = 0

Let:

cos (x) = u

- 5 u^{2} + 4 u + 1 = 0

- 5 u^{2} + 4 u + 1 = 0 : u = - \frac{1}{5}, u = 1

- 5 u^{2} + 4 u + 1 = 0

Solve with the quadratic formula

- 5 u^{2} + 4 u + 1 = 0

Quadratic Equation Formula:

For

a = - 5, b = 4, c = 1

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

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

4^{2} - 4 (- 5) \cdot 1 = 6

4^{2} - 4 (- 5) \cdot 1

Apply rule

- (- a) = a

= 4^{2} + 4 \cdot 5 \cdot 1

Multiply the numbers:

4 \cdot 5 \cdot 1 = 20

= 4^{2} + 20

4^{2} = 16

= 16 + 20

Add the numbers:

16 + 20 = 36

= 36

Factor the number:

36 = 6^{2}

= 6^{2}

Apply radical rule:

n a^{n} = a

6^{2} = 6

= 6

u_{1, 2} = \frac{- 4 \pm 6}{2 ( - 5 )}

Separate the solutions

u_{1} = \frac{- 4 + 6}{2 ( - 5 )}, u_{2} = \frac{- 4 - 6}{2 ( - 5 )}

u = \frac{- 4 + 6}{2 ( - 5 )} : - \frac{1}{5}

\frac{- 4 + 6}{2 ( - 5 )}

Remove parentheses:

(- a) = - a

= \frac{- 4 + 6}{- 2 \cdot 5}

Add/Subtract the numbers:

- 4 + 6 = 2

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

Multiply the numbers:

2 \cdot 5 = 10

= \frac{2}{- 10}

Apply the fraction rule:

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

= - \frac{2}{10}

Cancel the common factor:

2

= - \frac{1}{5}

u = \frac{- 4 - 6}{2 ( - 5 )} : 1

\frac{- 4 - 6}{2 ( - 5 )}

Remove parentheses:

(- a) = - a

= \frac{- 4 - 6}{- 2 \cdot 5}

Subtract the numbers:

- 4 - 6 = - 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{1}{5}, u = 1

Substitute back

u = cos (x)

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

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

cos (x) = - \frac{1}{5} : x = arccos (- \frac{1}{5}) + 2 πn, x = - arccos (- \frac{1}{5}) + 2 πn

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

Apply trig inverse properties

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

General solutions for

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

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

x = arccos (- \frac{1}{5}) + 2 πn, x = - arccos (- \frac{1}{5}) + 2 πn

x = arccos (- \frac{1}{5}) + 2 πn, x = - arccos (- \frac{1}{5}) + 2 πn

cos (x) = 1 : x = 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 = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

Combine all the solutions

x = arccos (- \frac{1}{5}) + 2 πn, x = - arccos (- \frac{1}{5}) + 2 πn, x = 2 πn

Show solutions in decimal form

x = 1.77215 \dots + 2 πn, x = - 1.77215 \dots + 2 πn, x = 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for -5cos^2(x)+4cos(x)+1=0 ?
The general solution for -5cos^2(x)+4cos(x)+1=0 is x=1.77215…+2pin,x=-1.77215…+2pin,x=2pin