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 $$\tan x<0$$?

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.

• Thats wrong, s = -1/2 is 2kpi + 7pi/6 and 2kpi - pi/6 , out of which only 2kpi - pi/6 is valid,
Feb 15, 2023 at 17:16
• Thanks @ Sinha, you are right. I corrected it, it should be only in the fourth quadrant. Feb 15, 2023 at 20:37

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.

$$tanx < 0$$ 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 $$x = 2n\pi -\frac{\pi}{6}$$

$$\frac\pi2+2k\pi$$ 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 $$<0$$).

• I am not sure I see your point. The solutions of the equation $\sin x=a, |a|\le1$ are $\begin{cases}x=\alpha+2k\pi\\x=\pi-\alpha+2k\pi\end{cases}$ where $\alpha\in\left[-\dfrac{\pi}{2};\dfrac{\pi}{2}\right]:\sin\alpha=a.$ In our problem $\alpha=-\dfrac{\pi}{6}$. I have just made a typo. Feb 15, 2023 at 17:19
• @yinivem462 Yes, you "have just made a typo", that is what I said: you wrote $\frac{5\pi}6$ instead of $-\frac{5\pi}6$ (which is the same mod $2\pi$ as the new value $\frac{7\pi}6$ by which you replaced your previous $\frac{5\pi}6$). Feb 15, 2023 at 22:09