Question

Find all solutions to the equation

Answer 1

Answer:

$x=0, \\x=\pi$

Step-by-step explanation:

Recall the trigonometric identity $\sin 2x=2\sin x\cos x$.

Therefore, given $\sin 2x=2\sin x$, rewrite the left side of the equation:

$2\sin x\cos x=2\sin x$

Subtract $\sin x$ from both sides:

$2\sin x\cos x-2\sin x=0$

Factor out $2\sin x$ from both terms on the left:

$2\sin x(\cos x-1)=0$

We now have two cases:

$\begin{cases}2\sin x=0, x=k\pi\text{ for }k\in \mathbb{Z}\\\cos x-1=0,x=k2\pi\text{ for }k\in \mathbb{Z}\end{cases}$

Since the problem stipulates that $x$ is in the interval $[0, 2\pi)$, we have:

$\text{For }x\in [0, 2\pi):\\\sin x=0\rightarrow \boxed{x=0, x=\pi}\\\cos x-1=0\rightarrow \boxed{x=0}$

Recall that square brackets mean inclusive and parentheses mean exclusive. Therefore, $2\pi \notin [0, 2\pi)$.