What is the general solution for 6cos^2(x)-13cos(x)+6=0 ?

The general solution for 6cos^2(x)-13cos(x)+6=0 is x=0.84106…+2pin,x=2pi-0.84106…+2pin

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Popular >

6cos^2(x)-13cos(x)+6=0

Solution

6 cos^{2} (x) - 13 cos (x) + 6 = 0

Solution

x = 0.84106 \dots + 2 πn, x = 2 π - 0.84106 \dots + 2 πn

Degrees

x = 48.18968 \dots^{\circ} + 36 0^{\circ} n, x = 311.81031 \dots^{\circ} + 36 0^{\circ} n

Solution steps

6 cos^{2} (x) - 13 cos (x) + 6 = 0

Solve by substitution

6 cos^{2} (x) - 13 cos (x) + 6 = 0

Let:

cos (x) = u

6 u^{2} - 13 u + 6 = 0

6 u^{2} - 13 u + 6 = 0 : u = \frac{3}{2}, u = \frac{2}{3}

6 u^{2} - 13 u + 6 = 0

Solve with the quadratic formula

6 u^{2} - 13 u + 6 = 0

Quadratic Equation Formula:

For

a = 6, b = - 13, c = 6

u_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 6 \cdot 6}{2 \cdot 6}

u_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 6 \cdot 6}{2 \cdot 6}

(- 13)^{2} - 4 \cdot 6 \cdot 6 = 5

(- 13)^{2} - 4 \cdot 6 \cdot 6

Apply exponent rule:

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

n

is even

(- 13)^{2} = 1 3^{2}

= 1 3^{2} - 4 \cdot 6 \cdot 6

Multiply the numbers:

4 \cdot 6 \cdot 6 = 144

= 1 3^{2} - 144

1 3^{2} = 169

= 169 - 144

Subtract the numbers:

169 - 144 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

n a^{n} = a

5^{2} = 5

= 5

u_{1, 2} = \frac{- ( - 13 ) \pm 5}{2 \cdot 6}

Separate the solutions

u_{1} = \frac{- ( - 13 ) + 5}{2 \cdot 6}, u_{2} = \frac{- ( - 13 ) - 5}{2 \cdot 6}

u = \frac{- ( - 13 ) + 5}{2 \cdot 6} : \frac{3}{2}

\frac{- ( - 13 ) + 5}{2 \cdot 6}

Apply rule

- (- a) = a

= \frac{13 + 5}{2 \cdot 6}

Add the numbers:

13 + 5 = 18

= \frac{18}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{18}{12}

Cancel the common factor:

6

= \frac{3}{2}

u = \frac{- ( - 13 ) - 5}{2 \cdot 6} : \frac{2}{3}

\frac{- ( - 13 ) - 5}{2 \cdot 6}

Apply rule

- (- a) = a

= \frac{13 - 5}{2 \cdot 6}

Subtract the numbers:

13 - 5 = 8

= \frac{8}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{8}{12}

Cancel the common factor:

4

= \frac{2}{3}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

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

cos (x) = \frac{3}{2} :

No Solution

cos (x) = \frac{3}{2}

- 1 \leq cos (x) \leq 1

No Solution

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

cos (x) = \frac{2}{3}

Apply trig inverse properties

cos (x) = \frac{2}{3}

General solutions for

cos (x) = \frac{2}{3}

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

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

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

Combine all the solutions

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

Show solutions in decimal form

x = 0.84106 \dots + 2 πn, x = 2 π - 0.84106 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 6cos^2(x)-13cos(x)+6=0 ?
The general solution for 6cos^2(x)-13cos(x)+6=0 is x=0.84106…+2pin,x=2pi-0.84106…+2pin