# Is there always exactly one solution to $a \cos\left(\frac{x}{2}\right)- b \sin\left(\frac{x}{2}\right) = 0$ in the interval $0\leq x\leq 2\pi$?

I have a probably simple question. If I have an equation like \begin{align} a \cos \left(\frac{x}{2}\right) - b \sin \left(\frac{x}{2}\right) &= 0\\ \frac ab \cos \left(\frac{x}{2}\right) - \sin \left(\frac{x}{2}\right) &= 0 \end{align} with $$a, b > 0$$. Is it true that there is always precisely one solution in the interval $$0 \leq x \leq 2 \pi$$.

I think it is obvious, but can I use this in a proof without proving it?

• Never assume something is obvious if you cannot state why it is obvious. – Algebra geek Jan 19 '19 at 15:19

It is $$\tan(\frac{x}{2})=\frac{a}{b}$$
Simply note the definition of tangent: $$\tan(x) = \dfrac{\sin(x)}{\cos(x)}$$.
$$a\cos\left(\frac{x}{2}\right)-b\sin\left(\frac{x}{2}\right) = 0 \iff a\cos\left(\frac{x}{2}\right) = b\sin\left(\frac{x}{2}\right) \iff \\ \frac{a}{b} = \frac{\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)} \iff \frac{a}{b} = \tan\left(\frac{x}{2}\right)$$
Since the range of $$\tan(x)$$ includes all real numbers, there is a solution in the interval $$x \in [0, 2\pi]$$.