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Popular >

12sin^2(x)-cos(x)-6=0

Solution

12 sin^{2} (x) - cos (x) - 6 = 0

Solution

x = 2.41885 \dots + 2 πn, x = - 2.41885 \dots + 2 πn, x = 0.84106 \dots + 2 πn, x = 2 π - 0.84106 \dots + 2 πn

Degrees

x = 138.59037 \dots^{\circ} + 36 0^{\circ} n, x = - 138.59037 \dots^{\circ} + 36 0^{\circ} n, x = 48.18968 \dots^{\circ} + 36 0^{\circ} n, x = 311.81031 \dots^{\circ} + 36 0^{\circ} n

Solution steps

12 sin^{2} (x) - cos (x) - 6 = 0

Rewrite using trig identities

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

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

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

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

Simplify

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

a = 12, b = 1, c = cos^{2} (x)

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

Multiply the numbers:

12 \cdot 1 = 12

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

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

Simplify

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

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

Group like terms

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

Add/Subtract the numbers:

- 6 + 12 = 6

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 12, b = - 1, c = 6

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 12 ) \cdot 6}{2 ( - 12 )}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 12 ) \cdot 6}{2 ( - 12 )}

(- 1)^{2} - 4 (- 12) \cdot 6 = 17

(- 1)^{2} - 4 (- 12) \cdot 6

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 12 \cdot 6

(- 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 6 = 288

4 \cdot 12 \cdot 6

Multiply the numbers:

4 \cdot 12 \cdot 6 = 288

= 288

= 1 + 288

Add the numbers:

1 + 288 = 289

= 289

Factor the number:

289 = 1 7^{2}

= 1 7^{2}

Apply radical rule:

1 7^{2} = 17

= 17

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

Separate the solutions

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

u = \frac{- ( - 1 ) + 17}{2 ( - 12 )} : - \frac{3}{4}

\frac{- ( - 1 ) + 17}{2 ( - 12 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{1 + 17}{- 2 \cdot 12}

Add the numbers:

1 + 17 = 18

= \frac{18}{- 2 \cdot 12}

Multiply the numbers:

2 \cdot 12 = 24

= \frac{18}{- 24}

Apply the fraction rule:

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

= - \frac{18}{24}

Cancel the common factor:

6

= - \frac{3}{4}

u = \frac{- ( - 1 ) - 17}{2 ( - 12 )} : \frac{2}{3}

\frac{- ( - 1 ) - 17}{2 ( - 12 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{1 - 17}{- 2 \cdot 12}

Subtract the numbers:

1 - 17 = - 16

= \frac{- 16}{- 2 \cdot 12}

Multiply the numbers:

2 \cdot 12 = 24

= \frac{- 16}{- 24}

Apply the fraction rule:

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

= \frac{16}{24}

Cancel the common factor:

8

= \frac{2}{3}

The solutions to the quadratic equation are:

u = - \frac{3}{4}, u = \frac{2}{3}

Substitute back

u = cos (x)

cos (x) = - \frac{3}{4}, cos (x) = \frac{2}{3}

cos (x) = - \frac{3}{4}, cos (x) = \frac{2}{3}

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

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

Apply trig inverse properties

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

General solutions for

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

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

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

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

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

Show solutions in decimal form

x = 2.41885 \dots + 2 πn, x = - 2.41885 \dots + 2 πn, x = 0.84106 \dots + 2 πn, x = 2 π - 0.84106 \dots + 2 πn

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