Solve $\sin x+\cos x+\tan x+\csc x+\sec x+\cot x=-2$ in the interval $0
Solve $$\sin x+\cos x+\tan x+\csc x+\sec x+\cot x=-2$$ from $0<x<2\pi$.
Could you find the most elegant solution? Does it factorize?
 A: Substitute $\sin x + \cos x=t$
Also, $t^2=1+\sin 2x$
Given reduces to : 
$$\sin x+\cos x + \frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}+\frac{1}{\sin x}+\frac{1}{\cos x}$$
$$t+\frac{2}{t^2-1}+\frac{2t}{t^2-1}$$
$$t^3-t+2+2t=-2t^2+2$$
$$t^3+2t^2+t=0$$
$$t(t+1)^2=0$$
This is easy peasy.
Now you can easily solve the equation.
Then it's all about solving a nice trigonometry equation. You can use:$\frac{t}{\sqrt 2}=\sin(\frac{\pi}{4}+x)$
A: This might be considered an extended comment of lab bhattacharjee's answer above. 
The roots of the factored form are 
$(\cos x + \sin x)(\cos x + 1)(\sin x +1) = 0$
which gives
$\cos x = -\sin x$. The 2 solutions are $\dfrac{3\pi}{4}$ and $\dfrac{7\pi}{4}$;
$\cos x = -1$. The solution is $\pi$;
$\sin x = -1$. The solution is $\dfrac{3\pi}{2}$.
So on the interval $0 < x < 2\pi$, you have the solutions $\dfrac{3\pi}{4}, \pi, \dfrac{3\pi}{2}, \dfrac{7\pi}{4}$.
EDIT: Thanks to lab bhattacharjee for noticing which values will be undefined if evaluated in the original equation. The the only solutions that work are $\dfrac{3\pi}{4}$ and $\dfrac{7\pi}{4}$.
A: HINT:
Writing $c=\cos x,s=\sin x$
Multiply throughout by $cs$  $$c^2s+cs^2+c^2+s^2+c+s=-2cs\iff cs(c+s)+(c+s)^2+c+s=0$$
Now,
$$cs(c+s)+(c+s)^2+c+s=(c+s)(cs+c+s+1)=(c+s)(c+1)(s+1)$$
