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

The general solution for -5cos^2(θ)-4cos(θ)+4=0 is θ=0.95231…+2pin,θ=2pi-0.95231…+2pin

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Popular >

-5cos^2(θ)-4cos(θ)+4=0

Solution

- 5 cos^{2} (θ) - 4 cos (θ) + 4 = 0

Solution

θ = 0.95231 \dots + 2 πn, θ = 2 π - 0.95231 \dots + 2 πn

Degrees

θ = 54.56381 \dots^{\circ} + 36 0^{\circ} n, θ = 305.43618 \dots^{\circ} + 36 0^{\circ} n

Solution steps

- 5 cos^{2} (θ) - 4 cos (θ) + 4 = 0

Solve by substitution

- 5 cos^{2} (θ) - 4 cos (θ) + 4 = 0

Let:

cos (θ) = u

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

- 5 u^{2} - 4 u + 4 = 0 : u = - \frac{2 ( 1 + 6 )}{5}, u = \frac{2 ( 6 - 1 )}{5}

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

(- 4)^{2} - 4 (- 5) \cdot 4 = 46

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

Apply rule

- (- a) = a

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

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 4)^{2} = 4^{2}

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

Multiply the numbers:

4 \cdot 5 \cdot 4 = 80

= 4^{2} + 80

4^{2} = 16

= 16 + 80

Add the numbers:

16 + 80 = 96

= 96

Prime factorization of

96 : 2^{5} \cdot 3

96

96

divides by

296 = 48 \cdot 2

= 2 \cdot 48

48

divides by

248 = 24 \cdot 2

= 2 \cdot 2 \cdot 24

24

divides by

224 = 12 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3

= 2^{5} \cdot 3

= 2^{5} \cdot 3

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

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

Apply radical rule:

= 2^{4} 2 \cdot 3

Apply radical rule:

2^{4} = 2^{\frac{4}{2}} = 2^{2}

= 2^{2} 2 \cdot 3

Refine

= 46

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

Separate the solutions

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

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

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

Remove parentheses:

(- a) = - a, - (- a) = a

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

Multiply the numbers:

2 \cdot 5 = 10

= \frac{4 + 4 6}{- 10}

Apply the fraction rule:

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

= - \frac{4 + 4 6}{10}

Cancel

\frac{4 + 4 6}{10} : \frac{2 ( 1 + 6 )}{5}

\frac{4 + 4 6}{10}

Factor

4 + 46 : 4 (1 + 6)

4 + 46

Rewrite as

= 4 \cdot 1 + 46

Factor out common term

4

= 4 (1 + 6)

= \frac{4 ( 1 + 6 )}{10}

Cancel the common factor:

2

= \frac{2 ( 1 + 6 )}{5}

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

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

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

Remove parentheses:

(- a) = - a, - (- a) = a

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

Multiply the numbers:

2 \cdot 5 = 10

= \frac{4 - 4 6}{- 10}

Apply the fraction rule:

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

4 - 46 = - (46 - 4)

= \frac{4 6 - 4}{10}

Factor

46 - 4 : 4 (6 - 1)

46 - 4

Rewrite as

= 46 - 4 \cdot 1

Factor out common term

4

= 4 (6 - 1)

= \frac{4 ( 6 - 1 )}{10}

Cancel the common factor:

2

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

The solutions to the quadratic equation are:

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

Substitute back

u = cos (θ)

cos (θ) = - \frac{2 ( 1 + 6 )}{5}, cos (θ) = \frac{2 ( 6 - 1 )}{5}

cos (θ) = - \frac{2 ( 1 + 6 )}{5}, cos (θ) = \frac{2 ( 6 - 1 )}{5}

cos (θ) = - \frac{2 ( 1 + 6 )}{5} :

No Solution

cos (θ) = - \frac{2 ( 1 + 6 )}{5}

- 1 \leq cos (x) \leq 1

No Solution

cos (θ) = \frac{2 ( 6 - 1 )}{5} : θ = arccos (\frac{2 ( 6 - 1 )}{5}) + 2 πn, θ = 2 π - arccos (\frac{2 ( 6 - 1 )}{5}) + 2 πn

cos (θ) = \frac{2 ( 6 - 1 )}{5}

Apply trig inverse properties

cos (θ) = \frac{2 ( 6 - 1 )}{5}

General solutions for

cos (θ) = \frac{2 ( 6 - 1 )}{5}

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

θ = arccos (\frac{2 ( 6 - 1 )}{5}) + 2 πn, θ = 2 π - arccos (\frac{2 ( 6 - 1 )}{5}) + 2 πn

θ = arccos (\frac{2 ( 6 - 1 )}{5}) + 2 πn, θ = 2 π - arccos (\frac{2 ( 6 - 1 )}{5}) + 2 πn

Combine all the solutions

θ = arccos (\frac{2 ( 6 - 1 )}{5}) + 2 πn, θ = 2 π - arccos (\frac{2 ( 6 - 1 )}{5}) + 2 πn

Show solutions in decimal form

θ = 0.95231 \dots + 2 πn, θ = 2 π - 0.95231 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for -5cos^2(θ)-4cos(θ)+4=0 ?
The general solution for -5cos^2(θ)-4cos(θ)+4=0 is θ=0.95231…+2pin,θ=2pi-0.95231…+2pin