algebra-precalculustrigonometry
The question is to find the general form of:
$$\cot\theta=\tan8\theta$$
By using $$x=n\pi+\alpha$$
I obtained
$$n\pi+\frac{\pi}{2}-\theta=8\theta$$
and
$$n\pi+8\theta=\frac{\pi}{2}-\theta$$
Using this formula interchangeably yields two different results.
But only the first one is given as the answer, so why can't I use the formula interchangeably? Have I done something wrong?
In the picture below, the red line is all the points where $y=\frac12$. The green line is all the points where $y=\cos x$. Wherever the red and green lines intersect, we have both $y=\frac12$ and $y=\cos x$, so $\cos x = \frac12$.
Since the green line continues wiggling back and forth in the same manner, all the way out to infinity in both directions, and the red line continues straight to infinity in both directions, the red and green lines will intersect an infinite number of times; this shows that the equation $\cos x = \frac12$ has an infinite number of solutions.
Not all trigonometric equations have an infinite number of solutions. For example, $\cos x = 73$ has no solutions. $\cos x = kx$ has a finite number of solutions (except for $k=0$) whose number depends on $k$. For example, $\cos x = \frac1{10}x$ has 7 solutions.
Whenever you divide both sides of an equation by something, you are assuming that the thing you're dividing by is nonzero, because dividing by $0$ is not valid.
So going from $2 \sin \theta \cos \theta = \sin \theta$ to $2 \cos\theta = 1$ is only valid when $\sin\theta \ne 0$.
In general, all this means is that you need to check the $\sin \theta = 0$ case separately. For example, going from $2 \sin\theta \cos\theta = 1$ to $2 \cos\theta = \frac1{\sin\theta}$ is also only valid when $\sin\theta \ne 0$, but you don't lose any solutions by doing this, because values of $\theta$ for which $\sin\theta=0$ weren't solutions to begin with.
But in this particular case, when $\sin\theta = 0$, the equation $2\sin\theta \cos\theta = \sin\theta$ is satisfied, so it's correct to go from this to $$ 2\cos\theta = 1 \text{ or } \sin\theta = 0. $$
Best Answer
It's fine. Even though, the form of the answer is different, the set of solutions is the same. This is because $n$ can take negative values also.