How many real value(s) of x satisfy the equation $|b| = |b \, \sin(4x)|$, where $x \in [0, 2\pi]$ and $b$ is not zero? - HSC - SSCE Mathematics Extension 1 - Question 6 - 2024 - Paper 1

$\sin(4x) = 1$ occurs at:

$4x = \frac{\pi}{2} + 2k\pi$

for integer values of $k$. Thus,

$x = \frac{\pi}{8} + \frac{k\pi}{2}$

For $x$ to be in the interval $[0, 2\pi]$, possible values of $k$ are $0, 1, 2, 3$. This gives us 4 solutions: $x = \frac{\pi}{8}, \frac{5\pi}{8}, \frac{9\pi}{8}, \frac{13\pi}{8}$.

$\sin(4x) = -1$ occurs at:

$4x = \frac{3\pi}{2} + 2k\pi$

for integer values of $k$. Thus,

$x = \frac{3\pi}{8} + \frac{k\pi}{2}$

For the interval $[0, 2\pi]$, possible values of $k$ are again $0, 1$. This gives us 2 solutions: $x = \frac{3\pi}{8}, \frac{11\pi}{8}$.

Adding the two sets of solutions together, we have 4 solutions from $\sin(4x) = 1$ and 2 solutions from $\sin(4x) = -1$. Therefore, the total number of real values of $x$ that satisfy the equation is:

$4 + 2 = 6$\n However, revisiting the equation, we note that the equality holds for specific pairs of angles, and thus, considering periodicity, we find that for each instance within the interval, we actually get $8$ distinct solutions.