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Popular >

12cos^2(x)-cos(x)-1=0

Solution

12 cos^{2} (x) - cos (x) - 1 = 0

Solution

x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn, x = 1.82347 \dots + 2 πn, x = - 1.82347 \dots + 2 πn

Degrees

x = 70.52877 \dots^{\circ} + 36 0^{\circ} n, x = 289.47122 \dots^{\circ} + 36 0^{\circ} n, x = 104.47751 \dots^{\circ} + 36 0^{\circ} n, x = - 104.47751 \dots^{\circ} + 36 0^{\circ} n

Solution steps

12 cos^{2} (x) - cos (x) - 1 = 0

Solve by substitution

12 cos^{2} (x) - cos (x) - 1 = 0

Let:

cos (x) = u

12 u^{2} - u - 1 = 0

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

12 u^{2} - u - 1 = 0

Solve with the quadratic formula

12 u^{2} - u - 1 = 0

Quadratic Equation Formula:

For

a = 12, b = - 1, c = - 1

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

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

(- 1)^{2} - 4 \cdot 12 (- 1) = 7

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

Apply rule

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 12 \cdot 1 = 48

4 \cdot 12 \cdot 1

Multiply the numbers:

4 \cdot 12 \cdot 1 = 48

= 48

= 1 + 48

Add the numbers:

1 + 48 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- ( - 1 ) \pm 7}{2 \cdot 12}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 7}{2 \cdot 12}, u_{2} = \frac{- ( - 1 ) - 7}{2 \cdot 12}

u = \frac{- ( - 1 ) + 7}{2 \cdot 12} : \frac{1}{3}

\frac{- ( - 1 ) + 7}{2 \cdot 12}

Apply rule

- (- a) = a

= \frac{1 + 7}{2 \cdot 12}

Add the numbers:

1 + 7 = 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{- ( - 1 ) - 7}{2 \cdot 12} : - \frac{1}{4}

\frac{- ( - 1 ) - 7}{2 \cdot 12}

Apply rule

- (- a) = a

= \frac{1 - 7}{2 \cdot 12}

Subtract the numbers:

1 - 7 = - 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}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

cos (x) = \frac{1}{3}, cos (x) = - \frac{1}{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{1}{4} : x = arccos (- \frac{1}{4}) + 2 πn, x = - 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 = - arccos (- a) + 2 πn

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

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

Combine all the solutions

x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn, x = arccos (- \frac{1}{4}) + 2 πn, x = - arccos (- \frac{1}{4}) + 2 πn

Show solutions in decimal form

x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn, x = 1.82347 \dots + 2 πn, x = - 1.82347 \dots + 2 πn

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