Question

How do I solve: 2 sin (2x) - 2 sin x + 2√3 cos x - √3 = 0

Answer 1

Answer:

$\displaystyle x = \frac{\pi}{3} +k\, \pi$ or $\displaystyle x =- \frac{\pi}{3} +2\,k\, \pi$, where $k$ is an integer.

There are three such angles between $0$ and $2\pi$: $\displaystyle \frac{\pi}{3}$, $\displaystyle \frac{2\, \pi}{3}$, and $\displaystyle \frac{4\,\pi}{3}$.

Step-by-step explanation:

By the double angle identity of sines:

$\sin(2\, x) = 2\, \sin x \cdot \cos x$.

Rewrite the original equation with this identity:

$2\, (2\, \sin x \cdot \cos x) - 2\, \sin x + 2\sqrt{3}\, \cos x - \sqrt{3} = 0$.

Note, that $2\, (2\, \sin x \cdot \cos x)$ and $(-2\, \sin x)$ share the common factor $(2\, \sin x)$. On the other hand, $2\sqrt{3}\, \cos x$ and $(-\sqrt{3})$ share the common factor $\sqrt[3}$. Combine these terms pairwise using the two common factors:

$(2\, \sin x) \cdot (2\, \cos x - 1) + \left(\sqrt{3}\right)\, (2\, \cos x - 1) = 0$.

Note the new common factor $(2\, \cos x - 1)$. Therefore:

$\left(2\, \sin x + \sqrt{3}\right) \cdot (2\, \cos x - 1) = 0$.

This equation holds as long as either $\left(2\, \sin x + \sqrt{3}\right)$ or $(2\, \cos x - 1)$ is zero. Let $k$ be an integer. Accordingly:

• $\displaystyle \sin x = -\frac{\sqrt{3}}{2}$, which corresponds to $\displaystyle x = -\frac{\pi}{3} + 2\, k\, \pi$ and $\displaystyle x = -\frac{2\, \pi}{3} + 2\, k\, \pi$.
• $\displaystyle \cos x = \frac{1}{2}$, which corresponds to $\displaystyle x = \frac{\pi}{3} + 2\, k \, \pi$ and $\displaystyle x = -\frac{\pi}{3} + 2\, k \, \pi$.

Any $x$ that fits into at least one of these patterns will satisfy the equation. These pattern can be further combined:

• $\displaystyle x = \frac{\pi}{3} + k \, \pi$ (from $\displaystyle x = -\frac{2\,\pi}{3} + 2\, k\, \pi$ and $\displaystyle x = \frac{\pi}{3} + 2\, k \, \pi$, combined,) as well as
• $\displaystyle x =- \frac{\pi}{3} +2\,k\, \pi$.