cosx+sinx=sqrt(cosx)?

May 13, 2018

$\rightarrow x = 2 n \pi$ where $n \in \mathbb{Z}$

Explanation:

$\rightarrow \cos x + \sin x = \sqrt{\cos} x$

$\rightarrow \cos x - \sqrt{\cos} x = - \sin x$

$\rightarrow {\left(\cos x - \sqrt{\cos} x\right)}^{2} = {\left(- \sin x\right)}^{2}$

$\rightarrow {\cos}^{2} x - 2 \cos x \cdot \sqrt{\cos} x + \cos x = {\sin}^{2} x = 1 - {\cos}^{2} x$

$\rightarrow 2 {\cos}^{2} x - 2 \cos x \cdot \sqrt{\cos} x + \cos x - 1 = 0$

Let $\sqrt{\cos} x = y$ then $\cos x = {y}^{2}$

$\rightarrow 2 \cdot {\left({y}^{2}\right)}^{2} - 2 \cdot {y}^{2} \cdot y + {y}^{2} - 1 = 0$

$\rightarrow 2 {y}^{4} - 2 {y}^{3} + {y}^{2} - 1 = 0$

$\rightarrow 2 {y}^{3} \left(y - 1\right) + \left(y + 1\right) \cdot \left(y - 1\right) = 0$

$\rightarrow \left[y - 1\right] \left[2 {y}^{3} + y + 1\right] = 0$

Taking , $\rightarrow y - 1 = 0$

$\rightarrow \sqrt{\cos} x = 1$

$\rightarrow \cos x = 1 = \cos 0$

$\rightarrow x = 2 n \pi \pm 0 = 2 n \pi$ where $n \in \mathbb{Z}$ which is the general

solution for $x$.