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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 :)

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    $\begingroup$ Multiply through by $\cos x$... $\endgroup$ – abiessu Apr 20 '16 at 13:02
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$$\frac1{\cos(x)}-\cos(x)=\sin(x).$$

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

$$x=k\pi,\\x=\frac\pi4+k\pi.$$

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$$1=\cos^2x+\cos x\sin x\iff\cos x\sin x=\sin^2x$$

$$\iff\sin x(\cos x-\sin x)=0$$

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

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$$2=2\cos^2x+2\sin x\cos x=1+\cos2x+\sin2x$$

$$\cos2x+\sin2x=1\iff\cos\left(2x-\dfrac\pi4\right)=\cos\dfrac\pi4$$

$$\implies2x-\dfrac\pi4=2m\pi\pm\dfrac\pi4$$

where $m$ is an integer

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HINT:

$$1=\cos^2x+\cos x\sin x$$

As $\cos x\ne0,$ divide both sides by $\cos^2x$ to get

$$1+\tan^2x=1+\tan x$$

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Moving the $\cos$ to the LHS, the equation is equivalent to

$$\frac{\sin^2(x)}{\cos(x)}=\sin(x),$$i.e.

$$\sin(x)=0\text{ or }\sin(x)=\cos(x).$$

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