There is an exam question I tried to make up to test myself on trigonometry, but I am now confused...
An example of the style of question is:
The function f is defined such that $f(x) = 10 \cos x – 2 \sin x, 0 < x < 2 \pi$
(a) Express f(x) in the form $R\cos(x + \alpha)$, where $R$ and $\alpha$ are constants to be found
(b) Solve f(2x) = 4
The problem is with part (b) and not the computations... It is about what 'range' I should keep for my solutions.
Of course, $f(x)$ is defined on $0 < x < 2 \pi$ and so I have given 4 solutions at the moment, treating $f(2x)$ to also be defined on $0 < x < 2\pi$, because f is DEFINED on this domain. But then because we now have $2x$, the domain could change to $0 < 2x < 2\pi$, which gives $0< x< \pi$ and so I only need 2 of the solutions.
Which is the most natural domain to pick in this case? Is it convention or is there a 'correct' answer?
Suppose you try to plug in $x = \frac{3\pi}2$. Then $f(2x)=f(2\cdot \frac{3\pi}2)=f(3\pi)$ which is undefined because $3\pi$ is not in the domain of $f$.
Similar considerations make clear that only solutions in the domain $0 < x < \pi$ are admissible.