# How do i solve this question?

## $4 \sin 2 \left(x - 1\right) - 2 = 1$ between the domains of $- 2 \pi \le x \le 2 \pi$

#### Explanation:

Given that

$4 \setminus \sin 2 \left(x - 1\right) - 2 = 1$

$4 \setminus \sin 2 \left(x - 1\right) = 3$

$\setminus \sin 2 \left(x - 1\right) = \frac{3}{4}$

Writing general solutions,

$2 \left(x - 1\right) = 2 k \setminus \pi + \setminus {\sin}^{- 1} \left(\frac{3}{4}\right) \setminus \mathmr{and}$

$\setminus \setminus 2 \left(x - 1\right) = 2 k \setminus \pi + \setminus \pi - \setminus {\sin}^{- 1} \left(\frac{3}{4}\right)$

$x = k \setminus \pi + 1 + \frac{1}{2} \setminus {\sin}^{- 1} \left(\frac{3}{4}\right) \setminus \setminus \mathmr{and}$

$\setminus \setminus x = \left(2 k + 1\right) \setminus \frac{\pi}{2} + 1 - \frac{1}{2} \setminus {\sin}^{- 1} \left(\frac{3}{4}\right)$

Where, $k$ is any integer i.e. $k = 0 , \setminus \pm 1 , \setminus \pm 2 , \setminus \pm 3 , \setminus \ldots$

But it given that $- 2 \setminus \pi \setminus \le x \setminus \le 2 \setminus \pi$

hence setting $k = - 2 , - 1 , 0 , 1$ in above first general solution, we get four desired values

$x = - 2 \setminus \pi + 1 + \frac{1}{2} \setminus {\sin}^{- 1} \left(\frac{3}{4}\right) , \setminus - \setminus \pi + 1 + \frac{1}{2} \setminus {\sin}^{- 1} \left(\frac{3}{4}\right) , 1 + \frac{1}{2} \setminus {\sin}^{- 1} \left(\frac{3}{4}\right) \setminus \setminus , \setminus \pi + 1 + \frac{1}{2} \setminus {\sin}^{- 1} \left(\frac{3}{4}\right)$

or

$x = - 4.859 , - 1.717 , 1.424 , 4.565$

Similarly, setting $k = - 2 , - 1 , 0 , 1$ in above second general solution, we get another four desired values

$x = - \frac{3 \setminus \pi}{2} + 1 - \frac{1}{2} \setminus {\sin}^{- 1} \left(\frac{3}{4}\right) , - \frac{\setminus \pi}{2} + 1 - \frac{1}{2} \setminus {\sin}^{- 1} \left(\frac{3}{4}\right) , \frac{\setminus \pi}{2} + 1 - \frac{1}{2} \setminus {\sin}^{- 1} \left(\frac{3}{4}\right) \setminus \setminus , \frac{3 \setminus \pi}{2} + 1 - \frac{1}{2} \setminus {\sin}^{- 1} \left(\frac{3}{4}\right)$

or

$x = - 4.136 , - 0.995 , 2.147 , 5.288$

Thus we get total eight desired values of $x$ which are approximately

$x = - 4.859 , - 4.136 , - 1.717 , - 0.995 , 1.424 , 2.147 , 4.565 , 5.288$