Skip to main content
Lemma

TMUA practice: Trig Equations, problem 1

A TMUA-calibre trig equations problem at difficulty 4 of 5, with the full worked solution.

The question
How many solutions does the equation tan(2x)=cos(x2)\tan(2x)=\cos(\tfrac{x}{2}) have on the interval [0,2π]?[0,2\pi]?
A 11
B 22
C 33
D 44
E 55

The correct answer is highlighted. E

Worked solution

We count the intersections of the graphs of y=tan(2x)y=\tan(2x) and y=cos(x2):y=\cos\left(\frac x2\right):

  • The graph of y=tan(2x)y=\tan(2x) has a period of π2,\frac{\pi}{2}, asymptotes at x=π4+kπ2,x=\frac{\pi}{4}+\frac{k\pi}{2}, and zeros at x=kπ2x=\frac{k\pi}{2} for some integer k.k.

On the interval [0,2π],[0,2\pi], the graph has five branches:

[0,π4),(π4,3π4),(3π4,5π4),(5π4,7π4),(7π4,2π].\biggl[0,\frac{\pi}{4}\biggr),\left(\frac{\pi}{4},\frac{3\pi}{4}\right),\left(\frac{3\pi}{4},\frac{5\pi}{4}\right),\left(\frac{5\pi}{4},\frac{7\pi}{4}\right),\left(\frac{7\pi}{4},2\pi\right].

Note that tan(2x)[0,)\tan(2x)\in[0,\infty) for the first branch, tan(2x)(,)\tan(2x)\in(-\infty,\infty) for the three middle branches, and tan(2x)(,0]\tan(2x)\in(-\infty,0] for the last branch. Moreover, all branches are strictly increasing.

  • The graph of y=cos(x2)y=\cos\left(\frac x2\right) has a period of 4π4\pi and zeros at x=π+2kπx=\pi+2k\pi for some integer k.k.

On the interval [0,2π],[0,2\pi], note that cos(x2)[1,1].\cos\left(\frac x2\right)\in[-1,1]. Moreover, the graph is strictly decreasing.

The graphs of y=tan(2x)y=\tan(2x) and y=cos(x2)y=\cos\left(\frac x2\right) intersect once on each of the five branches of y=tan(2x),y=\tan(2x), as shown below:

Therefore, the answer is (E) 5.\boxed{\textbf{(E)}\ 5}.