What is the general solution for 3(2+cos(x))=5 ?

The general solution for 3(2+cos(x))=5 is x=1.91063…+2pin,x=-1.91063…+2pin

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3(2+cos(x))=5

3 (2 + cos (x)) = 5

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

Degrees

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

Solution steps

3 (2 + cos (x)) = 5

Divide both sides by

3

3 (2 + cos (x)) = 5

Divide both sides by

3

\frac{3 ( 2 + cos ( x ) )}{3} = \frac{5}{3}

Simplify

2 + cos (x) = \frac{5}{3}

2 + cos (x) = \frac{5}{3}

Move

2

to the right side

2 + cos (x) = \frac{5}{3}

Subtract

2

from both sides

2 + cos (x) - 2 = \frac{5}{3} - 2

Simplify

2 + cos (x) - 2 = \frac{5}{3} - 2

Simplify

2 + cos (x) - 2 : cos (x)

2 + cos (x) - 2

Add similar elements:

2 - 2 = 0

= cos (x)

Simplify

\frac{5}{3} - 2 : - \frac{1}{3}

\frac{5}{3} - 2

Convert element to fraction:

2 = \frac{2 \cdot 3}{3}

= - \frac{2 \cdot 3}{3} + \frac{5}{3}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- 2 \cdot 3 + 5}{3}

- 2 \cdot 3 + 5 = - 1

- 2 \cdot 3 + 5

Multiply the numbers:

2 \cdot 3 = 6

= - 6 + 5

Add/Subtract the numbers:

- 6 + 5 = - 1

= - 1

= \frac{- 1}{3}

Apply the fraction rule:

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

= - \frac{1}{3}

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

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

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

Show solutions in decimal form

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

What is the general solution for 3(2+cos(x))=5 ?
The general solution for 3(2+cos(x))=5 is x=1.91063…+2pin,x=-1.91063…+2pin