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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 $\tan x<0$?

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4 Answers 4

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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.

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

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

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

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  • $\begingroup$ 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. $\endgroup$
    – yinivem462
    Feb 15, 2023 at 17:19
  • $\begingroup$ @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 $\frac{5\pi}6$). $\endgroup$ Feb 15, 2023 at 22:09

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