Solve $\sin x+\cos x+\tan x+\csc x+\sec x+\cot x=-2$ in the interval $0<x<2\pi$

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, $$cs(c+s)+(c+s)^2+c+s=(c+s)(cs+c+s+1)=(c+s)(c+1)(s+1)$$

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:$\frac{t}{\sqrt 2}=\sin(\frac{\pi}{4}+x)$

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 $\dfrac{7\pi}{4}$.