Find all solutions of sin(2x) = cos(x) in [0, 2π]

Question

Find all solutions of $\sin(2x) = \cos(x)$ in the interval $[0, 2\pi]$.

Solution — Step by Step

Why This Works

The double angle identity $\sin(2x) = 2\sin x \cos x$ is essential here — it allows us to write the equation in terms of $\sin x$ and $\cos x$ at the same argument $x$, rather than the mixed $2x$ and $x$.

The factoring step is critical. The equation becomes $A \times B = 0$, which means $A = 0$ OR $B = 0$ — a fundamental property of real numbers (zero product property). Each factor is a simple trig equation on the interval $[0, 2\pi]$.

Alternative Method — Convert to cos only

$\sin(2x) = \cos(x)$

Use $\sin(2x) = \cos(\pi/2 - 2x)$ (co-function identity):

This gives two families:

1. $\frac{\pi}{2} - 2x = x + 2k\pi \Rightarrow 3x = \frac{\pi}{2} - 2k\pi$

   For $k = 0$: $x = \pi/6$. For $k = -1$: $x = \pi/6 + 2\pi/3 = 5\pi/6$. For $k = 1$: $x$ would be negative, out of range.

2. $\frac{\pi}{2} - 2x = -x + 2k\pi \Rightarrow -x = -\frac{\pi}{2} + 2k\pi \Rightarrow x = \frac{\pi}{2} - 2k\pi$

   For $k = 0$: $x = \pi/2$. For $k = -1$: $x = 5\pi/2$ (out of range). For $k = 1$: $x = -3\pi/2$ (out of range, but equivalent to $\pi/2$ modulo $2\pi$… checking: $\pi/2 - 2(-1)\pi = \pi/2 + 2\pi = 5\pi/2$, too large. Try $k=0$: $\pi/2$, and there is also $3\pi/2$).

   Actually: $x = \pi/2 - 2k\pi$. For $k = 0$: $\pi/2$. For $k = -1$: $x = \pi/2 + 2\pi = 5\pi/2$ (too large). But also the negative of the argument: $x = 2\pi - \pi/2 + 2k\pi = 3\pi/2$. So $x = 3\pi/2$.

The factoring method is simpler and less prone to error.

Common Mistake