To solve the trigonometric equation [tex]\(\sqrt{3} \sin x - 2 \sin x \cos x = 0\)[/tex], let's follow a detailed step-by-step method:
1. Factor the Common Term:
[tex]\[\sqrt{3} \sin x - 2 \sin x \cos x = 0\][/tex]
Notice that [tex]\(\sin x\)[/tex] is a common factor in both terms:
[tex]\[\sin x (\sqrt{3} - 2 \cos x) = 0\][/tex]
2. Set Each Factor to Zero:
For the product to be zero, either [tex]\(\sin x = 0\)[/tex] or [tex]\(\sqrt{3} - 2 \cos x = 0\)[/tex].
- Solving [tex]\(\sin x = 0\)[/tex]:
[tex]\[\sin x = 0\][/tex]
This happens at:
[tex]\[x = k\pi \quad (k \in \mathbb{Z})\][/tex]
- Solving [tex]\(\sqrt{3} - 2 \cos x = 0\)[/tex]:
[tex]\[\sqrt{3} = 2 \cos x\][/tex]
[tex]\[\cos x = \frac{\sqrt{3}}{2}\][/tex]
3. Find the Values of [tex]\(x\)[/tex] for [tex]\(\cos x = \frac{\sqrt{3}}{2}\)[/tex]:
Cosine equals [tex]\(\frac{\sqrt{3}}{2}\)[/tex] at:
[tex]\[x = \pm \frac{\pi}{6} + 2k\pi \quad (k \in \mathbb{Z})\][/tex]
4. Formalize the General Solutions:
- From [tex]\(\sin x = 0\)[/tex]:
[tex]\[x = 0, \pi, 2\pi, \ldots\][/tex]
- From [tex]\(\cos x = \frac{\sqrt{3}}{2}\)[/tex]:
[tex]\[x_1 = \frac{\pi}{6}, x_2 = -\frac{\pi}{6}, x_3 = \frac{5\pi}{6}, x_4 = -\frac{5\pi}{6} + 2k\pi \quad (k \in \mathbb{Z})\][/tex]
Using the numerical results, we see the specific solutions derived numerically as:
[tex]\[x = 0\][/tex]
[tex]\[x = -2\mathrm{atan}\left(\frac{\sqrt{2-\sqrt{3}}}{\sqrt{\sqrt{3}+2}}\right)\][/tex]
[tex]\[x = 2\mathrm{atan}\left(\frac{\sqrt{2-\sqrt{3}}}{\sqrt{\sqrt{3}+2}}\right)\][/tex]
Thus, the solutions to the equation [tex]\(\sqrt{3} \sin x - 2 \sin x \cos x = 0\)[/tex] are:
[tex]\[x = 0, -2\mathrm{atan}\left(\frac{\sqrt{2-\sqrt{3}}}{\sqrt{\sqrt{3}+2}}\right), 2\mathrm{atan}\left(\frac{\sqrt{2-\sqrt{3}}}{\sqrt{\sqrt{3}+2}}\right)\][/tex]
