Solve the equation $1 / \cos x = \cos x + \sin x$ I'm having trouble solving the equation
$$
\frac{1}{\cos x} = \cos x + \sin x
$$
For what I understand I have to make the equation $= 0$
So I get 
$$
\frac{1}{\cos x} - \cos x -\sin x = 0
$$
Any help is greatly appreciated :)
 A: $$2=2\cos^2x+2\sin x\cos x=1+\cos2x+\sin2x$$
$$\cos2x+\sin2x=1\iff\cos\left(2x-\dfrac\pi4\right)=\cos\dfrac\pi4$$
$$\implies2x-\dfrac\pi4=2m\pi\pm\dfrac\pi4$$
where $m$ is an integer
A: $$\frac1{\cos(x)}-\cos(x)=\sin(x).$$
Squaring and setting $\cos^2(x)=t$,
$$\frac1t-2+t=1-t.$$
The solutions of this second degree equation ($2t^2-3t+1=0$) are $t=1$ and $t=\frac12$, corresponding to
$$\cos(x)=\pm1\to x=k\pi,\\
\cos(x)=\pm\frac1{\sqrt2}\to\frac\pi4+\frac{k\pi}2.$$
Anyway, squaring the equation introduced extra solutions. After checking compatibility, we are left with
$$x=k\pi,\\x=\frac\pi4+k\pi.$$
A: $$1=\cos^2x+\cos x\sin x\iff\cos x\sin x=\sin^2x$$
$$\iff\sin x(\cos x-\sin x)=0$$
If $\sin x=0,x=n\pi$
Else $\cos x-\sin x=0\iff\tan x=1\implies x=m\pi+\dfrac\pi4$
where $m,n$ are integers
A: HINT:
$$1=\cos^2x+\cos x\sin x$$
As $\cos x\ne0,$ divide both sides by $\cos^2x$ to get
$$1+\tan^2x=1+\tan x$$
A: Moving the $\cos$ to the LHS, the equation is equivalent to
$$\frac{\sin^2(x)}{\cos(x)}=\sin(x),$$i.e.
$$\sin(x)=0\text{ or }\sin(x)=\cos(x).$$
