2013 Mock AIME I Problems/Problem 6
Problem 6
Find the number of integer values 
can have such that the equation ![\[7\cos x+5\sin x=2k+1\]](http://latex.artofproblemsolving.com/4/3/a/43a0679c9612e94f6b3c566f42ce670f09a1caa7.png)
has a solution.
Solution 1
is a continuous function, so every value between its minimum and maximum is attainable by the Intermediate Value Theorem. By Cauchy-Schwarz, ![\[(7\cos x+5\sin x)^2 \le (7^2+5^2)(\cos^2 x+\sin^2 x)=74\]](http://latex.artofproblemsolving.com/b/8/9/b89f8702b44ba86d019298cc245dfb71ad180bdd.png)
Giving a maximum of 
, which is achievable when 
. Note that a minimum of 
can be attained at 
. Thus the values of that work are the integers from 
to 
, inclusive, giving a total of 
.
Solution 2 (calculus)
As in the first solution, let . Then, 
. Thus, 
has maxima and minima when 
. After squaring both sides and applying the Pythagorean Identity, we solve for 
:
\begin{align*}
49\sin^2x &= 25\cos^2x \\
49\sin^2x &= 25(1-\sin^2x) \\
74\sin^2x &= 25 \\
\sin^2x &= \frac{25}{74} \\
\sin x &= \pm \frac5{\sqrt{74}}
\end{align*}
Thus, 
, so 
. The maximum of the function occurs when 
, so the maximum is 
. Likewise, when both and 
are negative, the minimum occurs. We can now proceed as in the first solution to solve the problem to get our answer .
