To solve the problem of finding the months when the temperature [tex]$f(x) = 5 \tan(\frac{x}{12})$[/tex] is [tex]$195^\circ C$[/tex], we follow these steps:
1. Set up the equation:
[tex]\[
f(x) = 5 \tan\left(\frac{x}{12}\right) = 195
\][/tex]
Simplify by dividing both sides of the equation by 5:
[tex]\[
\tan\left(\frac{x}{12}\right) = 39
\][/tex]
2. Solve for [tex]\( x \)[/tex]:
The general solution for [tex]\(\tan(\theta) = k\)[/tex] is given by:
[tex]\[
\theta = \arctan(k) + n\pi \quad \text{where } n \text{ is any integer}
\][/tex]
In this problem, [tex]\( \theta = \frac{x}{12} \)[/tex] and [tex]\( k = 39 \)[/tex]. So:
[tex]\[
\frac{x}{12} = \arctan(39) + n\pi
\][/tex]
3. Isolate [tex]\( x \)[/tex]:
Multiply through by 12:
[tex]\[
x = 12 \arctan(39) + 12n\pi
\][/tex]
4. Identify the periodicity:
Recognize that [tex]\( 12\pi \)[/tex] is equivalent to a full cycle of 24 months because [tex]$2\pi$[/tex] radians correspond to [tex]\(360^\circ\)[/tex], which equals a complete cycle in the context of months:
[tex]\[
x = 12 \arctan(39) + 24n
\][/tex]
Calculate the specific solutions by recognizing that [tex]\(\arctan(39)\)[/tex] produces a specific angle, but we need to consider the principal value and then the periodic solutions as follows:
For [tex]\( m \)[/tex] being an integer multiple corresponding to the periodicity, and including another principal value at [tex]\( \arctan(39) + \pi \)[/tex], which changes the cycle offsets.
5. General solution with integer multiples:
Finally, express the solution in relation to multiples of full cycles of months:
[tex]\[
x = 3m + 24mn \quad \text{ and } \quad x = 15m + 24mn
\][/tex]
Therefore, the months when the temperature will be [tex]\( 195^\circ C \)[/tex] can be represented by:
[tex]\[
x = 3m + 24mn \quad \text{and} \quad x = 15m + 24mn
\][/tex]
These equations represent the specific months within the cycles where the temperature will reach [tex]\( 195^\circ C \)[/tex].
