# Find the exact value? 2sinxcosx+sinx-2cosx=1

May 9, 2018

$\rightarrow x = 2 n \pi \pm \frac{2 \pi}{3}$ OR $x = n \pi + {\left(- 1\right)}^{n} \left(\frac{\pi}{2}\right)$ where $n \rightarrow Z$

#### Explanation:

$\rightarrow 2 \sin x \cdot \cos x + \sin x - 2 \cos x = 1$

$\rightarrow \sin x \left(2 \cos x + 1\right) - 2 \cos x - 1 =$

$\rightarrow \sin x \left(2 \cos x + 1\right) - 1 \left(2 \cos x + 1\right) = 0$

$\rightarrow \left(2 \cos x + 1\right) \left(\sin x - 1\right) = 0$

Either, $2 \cos x + 1 = 0$

$\rightarrow \cos x = - \frac{1}{2} = - \cos \left(\frac{\pi}{3}\right) = \cos \left(\pi - \frac{2 \pi}{3}\right) = \cos \left(\frac{2 \pi}{3}\right)$

$\rightarrow x = 2 n \pi \pm \frac{2 \pi}{3}$ where $n \rightarrow Z$

OR, $\sin x - 1 = 0$

$\rightarrow \sin x = 1 = \sin \left(\frac{\pi}{2}\right)$

$\rightarrow x = n \pi + {\left(- 1\right)}^{n} \left(\frac{\pi}{2}\right)$ where $n \rightarrow Z$