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Popular Trigonometry >

cos^2(x)+7/12 cos(x)+1/12 =0

Solution

cos^{2} (x) + \frac{7}{12} cos (x) + \frac{1}{12} = 0

Solution

x = 1.82347 \dots + 2 πn, x = - 1.82347 \dots + 2 πn, x = 1.91063 \dots + 2 πn, x = - 1.91063 \dots + 2 πn

Degrees

x = 104.47751 \dots^{\circ} + 36 0^{\circ} n, x = - 104.47751 \dots^{\circ} + 36 0^{\circ} n, x = 109.47122 \dots^{\circ} + 36 0^{\circ} n, x = - 109.47122 \dots^{\circ} + 36 0^{\circ} n

Solution steps

cos^{2} (x) + \frac{7}{12} cos (x) + \frac{1}{12} = 0

Solve by substitution

cos^{2} (x) + \frac{7}{12} cos (x) + \frac{1}{12} = 0

Let:

cos (x) = u

u^{2} + \frac{7}{12} u + \frac{1}{12} = 0

u^{2} + \frac{7}{12} u + \frac{1}{12} = 0 : u = - \frac{1}{4}, u = - \frac{1}{3}

u^{2} + \frac{7}{12} u + \frac{1}{12} = 0

Multiply both sides by

12

u^{2} + \frac{7}{12} u + \frac{1}{12} = 0

Multiply both sides by

12

u^{2} \cdot 12 + \frac{7}{12} u \cdot 12 + \frac{1}{12} \cdot 12 = 0 \cdot 12

Simplify

12 u^{2} + 7 u + 1 = 0

12 u^{2} + 7 u + 1 = 0

Solve with the quadratic formula

12 u^{2} + 7 u + 1 = 0

Quadratic Equation Formula:

For

a = 12, b = 7, c = 1

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 12 \cdot 1}{2 \cdot 12}

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 12 \cdot 1}{2 \cdot 12}

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

7^{2} - 4 \cdot 12 \cdot 1

Multiply the numbers:

4 \cdot 12 \cdot 1 = 48

= 7^{2} - 48

7^{2} = 49

= 49 - 48

Subtract the numbers:

49 - 48 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

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

Add/Subtract the numbers:

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

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

Subtract the numbers:

- 7 - 1 = - 8

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

Multiply the numbers:

2 \cdot 12 = 24

= \frac{- 8}{24}

Apply the fraction rule:

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

= - \frac{8}{24}

Cancel the common factor:

8

= - \frac{1}{3}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

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

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

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

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

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

Combine all the solutions

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

Show solutions in decimal form

x = 1.82347 \dots + 2 πn, x = - 1.82347 \dots + 2 πn, x = 1.91063 \dots + 2 πn, x = - 1.91063 \dots + 2 πn

Graph

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