# trig function solutions on a specific interval

How many solutions are there to the equation $\cos(31x) = \sqrt{3}/2$ on the interval $[-4\pi,2\pi)$? Justify your answer.

Since $\pi/6$ and $11\pi/6$ yields $\sqrt{3}/2$ in function $\cos(x)$, I did $31x = \pi/6+2n\pi$ and the same for the other and put both of them in $[-4\pi,2\pi)$ to solve the inequality. In the end, I got that (since n has to be a positive integer) $n=30$. However, the answer is supposed to be 186. How can I solve this?

• Note that $n$ can also be a negative integer. Also, the fact that $x \in [-4\pi, 2\pi)$ implies that $31x \in [-124\pi, 62\pi)$. Did you take this into account when counting the number of possible $n$'s? – levap Aug 13 '17 at 8:41

Solving $\cos(31x) = \frac{\sqrt{3}}{2}$ on $[-4\pi, 2\pi)$ is equivalent to solving $\cos(u) = \frac{\sqrt{3}}{2}$ on $[-124 \pi, 62 \pi)$. The reason is that as $x$ ranges in $[-4\pi, 2\pi)$, the variable $u = 31x$ ranges in $$[31 \cdot (-4\pi), 31 \cdot (2\pi)) = [-124\pi, 62 \pi).$$
On each interval of the form $[\pi k, \pi(k + 1))$ (where $k \in \mathbb{Z}$), the equation $\cos(u) = \frac{\sqrt{3}}{2}$ has only one solution. Hence, the total number of solutions is $124 + 62 = 186$.