What is the general solution for 12cos^2(x)+11cos(x)-5=0 ?

The general solution for 12cos^2(x)+11cos(x)-5=0 is x=1.23095…+2pin,x=2pi-1.23095…+2pin

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Popular >

12cos^2(x)+11cos(x)-5=0

Solution

12 cos^{2} (x) + 11 cos (x) - 5 = 0

Solution

x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn

Degrees

x = 70.52877 \dots^{\circ} + 36 0^{\circ} n, x = 289.47122 \dots^{\circ} + 36 0^{\circ} n

Solution steps

12 cos^{2} (x) + 11 cos (x) - 5 = 0

Solve by substitution

12 cos^{2} (x) + 11 cos (x) - 5 = 0

Let:

cos (x) = u

12 u^{2} + 11 u - 5 = 0

12 u^{2} + 11 u - 5 = 0 : u = \frac{1}{3}, u = - \frac{5}{4}

12 u^{2} + 11 u - 5 = 0

Solve with the quadratic formula

12 u^{2} + 11 u - 5 = 0

Quadratic Equation Formula:

For

a = 12, b = 11, c = - 5

u_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 12 ( - 5 )}{2 \cdot 12}

u_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 12 ( - 5 )}{2 \cdot 12}

1 1^{2} - 4 \cdot 12 (- 5) = 19

1 1^{2} - 4 \cdot 12 (- 5)

Apply rule

- (- a) = a

= 1 1^{2} + 4 \cdot 12 \cdot 5

Multiply the numbers:

4 \cdot 12 \cdot 5 = 240

= 1 1^{2} + 240

1 1^{2} = 121

= 121 + 240

Add the numbers:

121 + 240 = 361

= 361

Factor the number:

361 = 1 9^{2}

= 1 9^{2}

Apply radical rule:

1 9^{2} = 19

= 19

u_{1, 2} = \frac{- 11 \pm 19}{2 \cdot 12}

Separate the solutions

u_{1} = \frac{- 11 + 19}{2 \cdot 12}, u_{2} = \frac{- 11 - 19}{2 \cdot 12}

u = \frac{- 11 + 19}{2 \cdot 12} : \frac{1}{3}

\frac{- 11 + 19}{2 \cdot 12}

Add/Subtract the numbers:

- 11 + 19 = 8

= \frac{8}{2 \cdot 12}

Multiply the numbers:

2 \cdot 12 = 24

= \frac{8}{24}

Cancel the common factor:

8

= \frac{1}{3}

u = \frac{- 11 - 19}{2 \cdot 12} : - \frac{5}{4}

\frac{- 11 - 19}{2 \cdot 12}

Subtract the numbers:

- 11 - 19 = - 30

= \frac{- 30}{2 \cdot 12}

Multiply the numbers:

2 \cdot 12 = 24

= \frac{- 30}{24}

Apply the fraction rule:

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

= - \frac{30}{24}

Cancel the common factor:

6

= - \frac{5}{4}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

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

cos (x) = \frac{1}{3} : x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

cos (x) = \frac{1}{3}

Apply trig inverse properties

cos (x) = \frac{1}{3}

General solutions for

cos (x) = \frac{1}{3}

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

x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

cos (x) = - \frac{5}{4} :

No Solution

cos (x) = - \frac{5}{4}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

Show solutions in decimal form

x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn

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sin (b) = 0.4848

Frequently Asked Questions (FAQ)

What is the general solution for 12cos^2(x)+11cos(x)-5=0 ?
The general solution for 12cos^2(x)+11cos(x)-5=0 is x=1.23095…+2pin,x=2pi-1.23095…+2pin