I was given this problem in my pre-calculus class: "Solve this trig equation for $0 \leq x < \pi$ (All angles are in radians) Equation: $\sin(2x) - \cos(x) = 0$."
Here are the steps I took in my attempt to solve the problem:
- I used the trig identity $\sin(2x) = 2\sin(x)\cos(2x)$ to substitute $2\sin(x)\cos(2x)$ for $\sin(2x)$ into the equation to get $2\sin(x)\cos(x) - \cos(x) = 0$.
- I added $\cos(x)$ to both sides of the equation to get $2\sin(x)\cos(x) = \cos(x)$, then I divided both sides by $\cos(x)$ to get $2\sin(x) = 1$.
- Finally, I divided both sides by $2$ to get $\sin(x) = 1/2$. Since $\arcsin(1/2) = \pi/6$ and $5\pi/6$, I wrote those numbers down as the solutions.
However, my teacher said that there was a third solution, and that I should have factored $2\sin(x)\cos(x) - \cos(x)$ into $\cos(x)[2\sin(x) - 1]$, then set both factors equal to $0$ to get $2\sin(x) - 1 = 0$ and $\cos(x) = 0$. Apparently, the solution I was missing was $\arccos(0) = \pi/2$.
Now after all the background details, here is my question: why didn't my original method find just $2$ of the $3$ solutions? Why couldn't I just get rid of the $\cos(x)$ like I did in step 2?