What is the general solution for 16=4+9-12cos(x) ?

The general solution for 16=4+9-12cos(x) is x=1.82347…+2pin,x=-1.82347…+2pin

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16=4+9-12cos(x)

16 = 4 + 9 - 12 cos (x)

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

Degrees

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

Solution steps

16 = 4 + 9 - 12 cos (x)

Switch sides

4 + 9 - 12 cos (x) = 16

Add the numbers:

4 + 9 = 13

- 12 cos (x) + 13 = 16

Move

13

to the right side

- 12 cos (x) + 13 = 16

Subtract

13

from both sides

- 12 cos (x) + 13 - 13 = 16 - 13

Simplify

- 12 cos (x) = 3

- 12 cos (x) = 3

Divide both sides by

- 12

- 12 cos (x) = 3

Divide both sides by

- 12

\frac{- 12 cos ( x )}{- 12} = \frac{3}{- 12}

Simplify

\frac{- 12 cos ( x )}{- 12} = \frac{3}{- 12}

Simplify

\frac{- 12 cos ( x )}{- 12} : cos (x)

\frac{- 12 cos ( x )}{- 12}

Apply the fraction rule:

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

= \frac{12 cos ( x )}{12}

Divide the numbers:

\frac{12}{12} = 1

= cos (x)

Simplify

\frac{3}{- 12} : - \frac{1}{4}

\frac{3}{- 12}

Apply the fraction rule:

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

= - \frac{3}{12}

Cancel the common factor:

3

= - \frac{1}{4}

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

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

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

Show solutions in decimal form

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

tanh (z) + 2 = 0

cos^{2} (a) = \frac{2}{3 sin ( a )}

5 cos (5 x) = 2

4 sin (x) - sin^{3} (x) - 1 = 0

1 - tan (a) = \frac{- 1}{3}

What is the general solution for 16=4+9-12cos(x) ?
The general solution for 16=4+9-12cos(x) is x=1.82347…+2pin,x=-1.82347…+2pin