What is the general solution for 6-4cos^2(x)-5cos(x)=0 ?

The general solution for 6-4cos^2(x)-5cos(x)=0 is x=0.72273…+2pin,x=2pi-0.72273…+2pin

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6-4cos^2(x)-5cos(x)=0

Solution

6 - 4 cos^{2} (x) - 5 cos (x) = 0

Solution

x = 0.72273 \dots + 2 πn, x = 2 π - 0.72273 \dots + 2 πn

Degrees

x = 41.40962 \dots^{\circ} + 36 0^{\circ} n, x = 318.59037 \dots^{\circ} + 36 0^{\circ} n

Solution steps

6 - 4 cos^{2} (x) - 5 cos (x) = 0

Solve by substitution

6 - 4 cos^{2} (x) - 5 cos (x) = 0

Let:

cos (x) = u

6 - 4 u^{2} - 5 u = 0

6 - 4 u^{2} - 5 u = 0 : u = - 2, u = \frac{3}{4}

6 - 4 u^{2} - 5 u = 0

Write in the standard form

a x^{2} + b x + c = 0

- 4 u^{2} - 5 u + 6 = 0

Solve with the quadratic formula

- 4 u^{2} - 5 u + 6 = 0

Quadratic Equation Formula:

For

a = - 4, b = - 5, c = 6

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

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

(- 5)^{2} - 4 (- 4) \cdot 6 = 11

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

(- 5)^{2} = 5^{2}

= 5^{2} + 4 \cdot 4 \cdot 6

Multiply the numbers:

4 \cdot 4 \cdot 6 = 96

= 5^{2} + 96

5^{2} = 25

= 25 + 96

Add the numbers:

25 + 96 = 121

= 121

Factor the number:

121 = 1 1^{2}

= 1 1^{2}

Apply radical rule:

1 1^{2} = 11

= 11

u_{1, 2} = \frac{- ( - 5 ) \pm 11}{2 ( - 4 )}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 11}{2 ( - 4 )}, u_{2} = \frac{- ( - 5 ) - 11}{2 ( - 4 )}

u = \frac{- ( - 5 ) + 11}{2 ( - 4 )} : - 2

\frac{- ( - 5 ) + 11}{2 ( - 4 )}

Remove parentheses:

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

= \frac{5 + 11}{- 2 \cdot 4}

Add the numbers:

5 + 11 = 16

= \frac{16}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{16}{- 8}

Apply the fraction rule:

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

= - \frac{16}{8}

Divide the numbers:

\frac{16}{8} = 2

= - 2

u = \frac{- ( - 5 ) - 11}{2 ( - 4 )} : \frac{3}{4}

\frac{- ( - 5 ) - 11}{2 ( - 4 )}

Remove parentheses:

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

= \frac{5 - 11}{- 2 \cdot 4}

Subtract the numbers:

5 - 11 = - 6

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 6}{- 8}

Apply the fraction rule:

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

= \frac{6}{8}

Cancel the common factor:

2

= \frac{3}{4}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

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

cos (x) = - 2 :

No Solution

cos (x) = - 2

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = \frac{3}{4} : x = arccos (\frac{3}{4}) + 2 πn, x = 2 π - arccos (\frac{3}{4}) + 2 πn

cos (x) = \frac{3}{4}

Apply trig inverse properties

cos (x) = \frac{3}{4}

General solutions for

cos (x) = \frac{3}{4}

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

x = arccos (\frac{3}{4}) + 2 πn, x = 2 π - arccos (\frac{3}{4}) + 2 πn

x = arccos (\frac{3}{4}) + 2 πn, x = 2 π - arccos (\frac{3}{4}) + 2 πn

Combine all the solutions

x = arccos (\frac{3}{4}) + 2 πn, x = 2 π - arccos (\frac{3}{4}) + 2 πn

Show solutions in decimal form

x = 0.72273 \dots + 2 πn, x = 2 π - 0.72273 \dots + 2 πn

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tan (a) = \frac{51}{140}

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Frequently Asked Questions (FAQ)

What is the general solution for 6-4cos^2(x)-5cos(x)=0 ?
The general solution for 6-4cos^2(x)-5cos(x)=0 is x=0.72273…+2pin,x=2pi-0.72273…+2pin