Question

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

Answer 1

`\displaystyle x = (\pi)/(3) +k\, \pi` or
`\displaystyle x =- (\pi)/(3) +2\,k\, \pi`, where
`k` is an integer.

There are three such angles between
`0` and
`2\pi`:
`\displaystyle (\pi)/(3)`,
`\displaystyle (2\, \pi)/(3)`, and
`\displaystyle (4\,\pi)/(3)`.

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√(3)\, \cos x - √(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√(3)\, \cos x` and
`(-√(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(√(3)\right)\, (2\, \cos x - 1) = 0`.

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

`\left(2\, \sin x + √(3)\right) \cdot (2\, \cos x - 1) = 0`.

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

• 
  `\displaystyle \sin x = -(√(3))/(2)`, which corresponds to
  `\displaystyle x = -(\pi)/(3) + 2\, k\, \pi` and
  `\displaystyle x = -(2\, \pi)/(3) + 2\, k\, \pi`.
• 
  `\displaystyle \cos x = (1)/(2)`, which corresponds to
  `\displaystyle x = (\pi)/(3) + 2\, k \, \pi` and
  `\displaystyle x = -(\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 = (\pi)/(3) + k \, \pi` (from
  `\displaystyle x = -(2\,\pi)/(3) + 2\, k\, \pi` and
  `\displaystyle x = (\pi)/(3) + 2\, k \, \pi`, combined,) as well as
• 
  `\displaystyle x =- (\pi)/(3) +2\,k\, \pi`.