# How to solve this equation: secx - cosx = sinx? Thank you!

Feb 19, 2018

$\sec x - \cos x = \sin x$

$\frac{1}{\cos} x - \cos x = \sin x$

$\frac{1 - {\cos}^{2} x}{\cos} x = \sin x$

${\sin}^{2} \frac{x}{\cos} x = \sin x$

${\sin}^{2} \frac{x}{\cos} x - \sin x = 0$

$\left(\sin x\right) \left(\sin \frac{x}{\cos} x - 1\right) = 0$

Here, $\sin x = 0$ or

$\tan x = 1$

$x = {\sin}^{- 1} 0$ or $x = {\tan}^{- 1} 1$

Hence, the general solution would be {kπ}∪{π/4+kπ},k∈ZZ