Find the roots of the equation $(1+\tan^2x)\sin x-\tan^2x+1=0$ which satisfy the inequality $\tan x<0$ Find the roots of the equation $$(1+\tan^2x)\sin x-\tan^2x+1=0$$ which satisfy the inequality $$\tan x<0$$
Shold I solve the equation first and then try to find which of the roots satisfy the inequality? Should I use $\tan x$ in the solution itself and obtain only the needed roots?
I wasn't able to see how the inequality can be used beforehand, so I solved the equation. For $x\ne \dfrac{\pi}{2}(2k+1),k\in\mathbb{Z}$, the equation is $$\dfrac{\sin^2x+\cos^2x}{\cos^2x}\sin x-\left(\dfrac{\sin^2x}{\cos^2x}-\dfrac{\cos^2x}{\cos^2x}\right)=0$$ which is equivalent to $$\sin x+\cos2x=0\\\sin x+1-2\sin^2x=0\\2\sin^2x-\sin x-1=0$$ which gives for the sine $-\dfrac12$ or $1$. Then the solutions are $$\begin{cases}x=-\dfrac{\pi}{6}+2k\pi\\x=\dfrac{7\pi}{6}+2k\pi\end{cases}\cup x=\dfrac{\pi}{2}+2k\pi$$ How do I use $\tan x<0$?
 A: Let $\sin x = s$; the given equation reduces to:
$$2s^2-s-1=0,~(s-1)(2s+1)=0$$
we have in the first full rotation of the radius vector
$$s=1\to x= \frac{\pi}{2},~\frac{3 \pi}{2}$$
$$s=-\frac{1}{2}\to x= \frac{7 \pi}{6},~\frac{11 \pi}{6}$$
between which only $x= \dfrac{11 \pi}{6}$ has its tangent negative.
This and its co-terminals $ x= 11 \pi/6 +2 k \pi$ satisfy the given equation.
A: It seems like you forgot a minus sign in front of $\frac{5\pi}{6}$ according to @AnneBauval, so your third solution doesn't work (it gives $\sqrt{3}$). Your second solution isn't in the domain since as you said at the beginning that $x\ne\frac{\pi}{2}(2k+1)=\pi k+\frac{\pi}{2}$ and substituting $k\mapsto2k$ tells us that the second solution isn't allowed. So the first solution is the only one that is usable.
A: $\frac\pi2+2k\pi$ should never appear because the initial equation is not defined when $\cos x=0.$
You forgot a $-$ sign in front of $\frac{5\pi}6+2k\pi.$
As for your final question: simply discard $-\frac{5\pi}6+2k\pi$ (whose $\tan$ is $>0$) and retain $-\frac\pi6+2k\pi$ (whose $\tan$ is $<0$).
A: $tanx < 0 $ essentially means $ x \in \frac{(2n-1)\pi}{2} ,n\pi $ and the by taking intersection of this and the solution that you obtained, the final answer is $x = 2n\pi -\frac{\pi}{6}$
