Determine the exact roots of \(\sqrt{2}\sin(x)\cos(x)=\cos(x)\), where \(-\pi\leq x \leq 2\pi\).

Solution

First, we note that there are no restrictions on the variable \(x\), since sine and cosine are defined for all real values of \(x\).

Determine the exact roots of \(\sqrt{2}\sin(x)\cos(x)=\cos(x)\), where \(-\pi\leq x \leq 2\pi\).

Solution

Collect the terms to one side of the equation and factor,

\(\cos(x)=0\) or \(\sqrt{2}\sin(x)-1=0\)

We know \(\cos(x)=0\) when \(x=\frac{\pi}{2}+n\pi\), \(n \in \mathbb{Z}\).

Thus, for \(-\pi \leq x \leq 2\pi\), \(x=-\frac{\pi}{2}\), \(\frac{\pi}{2}\), and \(\frac{3\pi}{2}\).

We know \(\sqrt{2}\sin(x)-1=0\) when \(\sin(x)=\frac{1}{\sqrt{2}}\).

The reference angle is \(\frac{\pi}{4}\) and sine is positive for any angle with a terminal arm in quadrant 1 or quadrant 2.

For \(0\leq x \leq 2\pi\), \(x=\frac{\pi}{4}\) and \(x=\pi - \frac{\pi}{4} = \frac{3\pi}{4}\).

For \(-\pi \leq x \leq 0\), \(\sin(x) \leq 0\) so no additional solutions lie in this part of the required domain.

Thus, \(x=\frac{\pi}{4}\) and \(x=\frac{3\pi}{4}\).

Therefore, the exact roots of \(\sqrt{2}\sin(x)\cos(x)=\cos(x)\), where \(-\pi \leq x \leq 2\pi\) are \(-\frac{\pi}{2}\), \(\frac{\pi}{4}\), \(\frac{\pi}{2}\), \(\frac{3\pi}{4}\), and \(\frac{3\pi}{2}\).

Determine the exact roots of \(\sqrt{2}\sin(x)\cos(x)=\cos(x)\), where \(-\pi\leq x \leq 2\pi\).

Solution

One way to verify this solution graphically using technology, is to graph

and determine the zeros or \(x\)-intercepts of the function.

The graph presented here allows for a quick visual check of the solution.

The zeros occur when \(f(x)=0\); that is, when \(\sqrt{2}\sin(x)\cos(x)-\cos(x)=0\).

Zeros shown in the domain \(-\pi \leq x \leq 2\pi\) are \(-\dfrac{\pi}{2}\), \(\dfrac{\pi}{4}\), \(\dfrac{\pi}{2}\), \(\dfrac{3\pi}{4}\), and \(\dfrac{3\pi}{2}\).