To find all solutions of the equation [tex]\(\sin x(2 \cos x + 1) = 0\)[/tex] in the interval [tex]\([0, 2\pi)\)[/tex], we need to consider the two factors of the product separately and solve them individually.
### Step 1: Solve [tex]\(\sin x = 0\)[/tex]
The sine function is zero at:
[tex]\[ x = n\pi \][/tex]
within the interval [tex]\([0, 2\pi)\)[/tex], where [tex]\(n\)[/tex] is an integer. We focus on values within the given interval:
[tex]\[ x = 0 \][/tex]
[tex]\[ x = \pi \][/tex]
### Step 2: Solve [tex]\(2 \cos x + 1 = 0\)[/tex]
Solve for [tex]\(\cos x\)[/tex]:
[tex]\[ 2 \cos x + 1 = 0 \][/tex]
[tex]\[ \cos x = -\frac{1}{2} \][/tex]
The cosine function equals [tex]\(-\frac{1}{2}\)[/tex] at:
[tex]\[ x = \frac{2\pi}{3} + 2k\pi \][/tex]
[tex]\[ x = \frac{4\pi}{3} + 2k\pi \][/tex]
where [tex]\(k\)[/tex] is any integer. We select values within the interval [tex]\([0, 2\pi)\)[/tex]:
[tex]\[ x = \frac{2\pi}{3} \][/tex]
[tex]\[ x = \frac{4\pi}{3} \][/tex]
### Step 3: Combine the solutions
Gather all solutions found:
[tex]\[ x = 0, \pi, \frac{2\pi}{3}, \frac{4\pi}{3} \][/tex]
To summarize, the solutions to the equation [tex]\(\sin x (2 \cos x + 1) = 0\)[/tex] in the interval [tex]\([0, 2\pi)\)[/tex] are:
[tex]\[ x = 0, \frac{2\pi}{3}, \pi, \frac{4\pi}{3} \][/tex]
Hence, the solutions in radians are:
[tex]\[ x = 0, \frac{2\pi}{3}, \pi, \frac{4\pi}{3} \][/tex]
