Hello,
The question is: Solve sin(x) = cos(2x) for all x.
My solution:
sin(x) = cos(2x)
sin(x) = sin(1/2 pi - 2x)
x = 1/2 pi - 2x
3x = 1/2 pi
x = 1/6 pi
However, the question says "solve for all x". I understand that fact that there are multiple x-es since both functions are periodic. But how do I find the other x-es? Please help!
because you assumed that if sin(x) = sin(pi/2 - 2x), that x = pi/2 - 2x. but this is not true, sin(x) is not a 1-1 function.
for example, sin(pi/2 - 2x) = sin(5pi/2 - 2x) as well (leading to x = 5pi/6), and also cos(2x) = cos(-2x), so
we have sin(x) = sin(pi/2 + 2x) also (leading to x = -pi/2, and thus (by adding 2pi) x = 3pi/2).
it is customary to find the solutions "in one revolution" (either from -pi to pi, or 0 to 2pi), and then add 2kpi to the results.