What is the general solution for -6cos(2x)+23cos(x)-11=0 ?

The general solution for -6cos(2x)+23cos(x)-11=0 is x=1.31811…+2pin,x=2pi-1.31811…+2pin

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Popular >

-6cos(2x)+23cos(x)-11=0

Solution

- 6 cos (2 x) + 23 cos (x) - 11 = 0

Solution

x = 1.31811 \dots + 2 πn, x = 2 π - 1.31811 \dots + 2 πn

Degrees

x = 75.52248 \dots^{\circ} + 36 0^{\circ} n, x = 284.47751 \dots^{\circ} + 36 0^{\circ} n

Solution steps

- 6 cos (2 x) + 23 cos (x) - 11 = 0

Rewrite using trig identities

- 11 + 23 cos (x) - 6 cos (2 x)

Use the Double Angle identity:

cos (2 x) = 2 cos^{2} (x) - 1

= - 11 + 23 cos (x) - 6 (2 cos^{2} (x) - 1)

Simplify

- 11 + 23 cos (x) - 6 (2 cos^{2} (x) - 1) : 23 cos (x) - 12 cos^{2} (x) - 5

- 11 + 23 cos (x) - 6 (2 cos^{2} (x) - 1)

Expand

- 6 (2 cos^{2} (x) - 1) : - 12 cos^{2} (x) + 6

- 6 (2 cos^{2} (x) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = - 6, b = 2 cos^{2} (x), c = 1

= - 6 \cdot 2 cos^{2} (x) - (- 6) \cdot 1

Apply minus-plus rules

- (- a) = a

= - 6 \cdot 2 cos^{2} (x) + 6 \cdot 1

Simplify

- 6 \cdot 2 cos^{2} (x) + 6 \cdot 1 : - 12 cos^{2} (x) + 6

- 6 \cdot 2 cos^{2} (x) + 6 \cdot 1

Multiply the numbers:

6 \cdot 2 = 12

= - 12 cos^{2} (x) + 6 \cdot 1

Multiply the numbers:

6 \cdot 1 = 6

= - 12 cos^{2} (x) + 6

= - 12 cos^{2} (x) + 6

= - 11 + 23 cos (x) - 12 cos^{2} (x) + 6

Simplify

- 11 + 23 cos (x) - 12 cos^{2} (x) + 6 : 23 cos (x) - 12 cos^{2} (x) - 5

- 11 + 23 cos (x) - 12 cos^{2} (x) + 6

Group like terms

= 23 cos (x) - 12 cos^{2} (x) - 11 + 6

Add/Subtract the numbers:

- 11 + 6 = - 5

= 23 cos (x) - 12 cos^{2} (x) - 5

= 23 cos (x) - 12 cos^{2} (x) - 5

= 23 cos (x) - 12 cos^{2} (x) - 5

- 5 - 12 cos^{2} (x) + 23 cos (x) = 0

Solve by substitution

- 5 - 12 cos^{2} (x) + 23 cos (x) = 0

Let:

cos (x) = u

- 5 - 12 u^{2} + 23 u = 0

- 5 - 12 u^{2} + 23 u = 0 : u = \frac{1}{4}, u = \frac{5}{3}

- 5 - 12 u^{2} + 23 u = 0

Write in the standard form

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

- 12 u^{2} + 23 u - 5 = 0

Solve with the quadratic formula

- 12 u^{2} + 23 u - 5 = 0

Quadratic Equation Formula:

For

a = - 12, b = 23, c = - 5

u_{1, 2} = \frac{- 23 \pm 2 3 ^{2} - 4 ( - 12 ) ( - 5 )}{2 ( - 12 )}

u_{1, 2} = \frac{- 23 \pm 2 3 ^{2} - 4 ( - 12 ) ( - 5 )}{2 ( - 12 )}

2 3^{2} - 4 (- 12) (- 5) = 17

2 3^{2} - 4 (- 12) (- 5)

Apply rule

- (- a) = a

= 2 3^{2} - 4 \cdot 12 \cdot 5

Multiply the numbers:

4 \cdot 12 \cdot 5 = 240

= 2 3^{2} - 240

2 3^{2} = 529

= 529 - 240

Subtract the numbers:

529 - 240 = 289

= 289

Factor the number:

289 = 1 7^{2}

= 1 7^{2}

Apply radical rule:

n a^{n} = a

1 7^{2} = 17

= 17

u_{1, 2} = \frac{- 23 \pm 17}{2 ( - 12 )}

Separate the solutions

u_{1} = \frac{- 23 + 17}{2 ( - 12 )}, u_{2} = \frac{- 23 - 17}{2 ( - 12 )}

u = \frac{- 23 + 17}{2 ( - 12 )} : \frac{1}{4}

\frac{- 23 + 17}{2 ( - 12 )}

Remove parentheses:

(- a) = - a

= \frac{- 23 + 17}{- 2 \cdot 12}

Add/Subtract the numbers:

- 23 + 17 = - 6

= \frac{- 6}{- 2 \cdot 12}

Multiply the numbers:

2 \cdot 12 = 24

= \frac{- 6}{- 24}

Apply the fraction rule:

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

= \frac{6}{24}

Cancel the common factor:

6

= \frac{1}{4}

u = \frac{- 23 - 17}{2 ( - 12 )} : \frac{5}{3}

\frac{- 23 - 17}{2 ( - 12 )}

Remove parentheses:

(- a) = - a

= \frac{- 23 - 17}{- 2 \cdot 12}

Subtract the numbers:

- 23 - 17 = - 40

= \frac{- 40}{- 2 \cdot 12}

Multiply the numbers:

2 \cdot 12 = 24

= \frac{- 40}{- 24}

Apply the fraction rule:

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

= \frac{40}{24}

Cancel the common factor:

8

= \frac{5}{3}

The solutions to the quadratic equation are:

u = \frac{1}{4}, u = \frac{5}{3}

Substitute back

u = cos (x)

cos (x) = \frac{1}{4}, cos (x) = \frac{5}{3}

cos (x) = \frac{1}{4}, cos (x) = \frac{5}{3}

cos (x) = \frac{1}{4} : x = arccos (\frac{1}{4}) + 2 πn, x = 2 π - arccos (\frac{1}{4}) + 2 πn

cos (x) = \frac{1}{4}

Apply trig inverse properties

cos (x) = \frac{1}{4}

General solutions for

cos (x) = \frac{1}{4}

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

x = arccos (\frac{1}{4}) + 2 πn, x = 2 π - arccos (\frac{1}{4}) + 2 πn

x = arccos (\frac{1}{4}) + 2 πn, x = 2 π - arccos (\frac{1}{4}) + 2 πn

cos (x) = \frac{5}{3} :

No Solution

cos (x) = \frac{5}{3}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = arccos (\frac{1}{4}) + 2 πn, x = 2 π - arccos (\frac{1}{4}) + 2 πn

Show solutions in decimal form

x = 1.31811 \dots + 2 πn, x = 2 π - 1.31811 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for -6cos(2x)+23cos(x)-11=0 ?
The general solution for -6cos(2x)+23cos(x)-11=0 is x=1.31811…+2pin,x=2pi-1.31811…+2pin