Question

Find all solutions in the interval [0,2л).

sin 2x cos x + sinx = 0

x =

Answer 1

Answer:

$x=0; x= \pi$

Step-by-step explanation:

Let's first convert that $sin\ 2x = 2 sinx \ cosx$ using the double angle formula.

$2 sinx\ cos^2x + sinx = 0$

Now, if $sin x = 0 \implies x= k\pi$ with $k\in \{0, 1\}$ we get $0=0$ which is indeed true, so we have a solution. Else, we can divide by $sin x \ne 0$ to get $cos^2x +1 = 0$ which has no real solutions (I like to think about it as the sum of two positive numbers, which is never zero).