Find all $x$ in the interval $(0,\pi/2)$ such that $\frac{\sqrt{3}-1}{\sin x}+\frac{\sqrt{3}+1}{\cos x}=4\sqrt{2}$.

The options are (i)$\pi/9,2\pi/7$, (ii)$\pi/36,11\pi/12$ (iii)$\pi/12,11\pi/36$ (iv) All

I have been able to find one value of $x$, $\pi/12$. How do I find the other root(s)?

My attempt:

$\frac{\sqrt{3}-1}{\sin x}+\frac{\sqrt{3}+1}{\cos x}=4\sqrt{2}$

or, $\frac{\sin\pi/3-\sin\pi/6}{\sin x}+\frac{\cos\pi/6+\cos\pi/3}{\cos x}=2\sqrt{2}$

or, $\frac{\sin(\pi/4)cos(\pi/12)}{\sin x}+\frac{\cos(\pi/4)cos(\pi/12)}{\cos x}=\sqrt{2}$

or, $\sin(x+\pi/12)=\sin2x$

or, $x=\pi/12$

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    $\begingroup$ How did you find the one value that you found? (Perhaps show your working -- it's possible that you accidentally lost some solutions.) $\endgroup$ – Minus One-Twelfth Mar 10 '19 at 11:53
  • $\begingroup$ I expressed the first numerator in terms of sine and the second in terms of cosine function. Then I used addition formula to combine both sines and cosines. After simplifying, I got $x=\pi/12$ $\endgroup$ – MrAP Mar 10 '19 at 11:56
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    $\begingroup$ Maybe you accidentally missed some solutions when you got to the simplified stage. Remember that if $\sin a = \sin b$, then all the solutions (plural!) or $a$ in terms of $b$ are $a = b + 2n\pi$ or $a = \pi - b + 2n\pi$, for some integer $n$. You will want to take only those solutions for which $x$ comes out to be between $0$ and $\pi/2$. $\endgroup$ – Minus One-Twelfth Mar 10 '19 at 11:58
  • $\begingroup$ @MinusOne-Twelfth, Check my question now. $\endgroup$ – MrAP Mar 10 '19 at 12:06
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    $\begingroup$ You have simply equated the angles. As I said in my previous comment, there are more solutions to $\sin a = \sin b$ than just $a=b$. This is why you were unable to obtain all the solutions. $\endgroup$ – Minus One-Twelfth Mar 10 '19 at 12:08

Be careful that your final equation has more potential solutions. The equation $$ \sin \left(x + \frac{\pi}{12}\right) = \sin 2x$$ implies in fact $$ x + \frac{\pi}{12} = 2x + 2k \pi$$ or $$ x + \frac{\pi}{12} = \pi - 2x + 2k \pi.$$

Also recall that you can always check the number of solutions by intersecting $$ \frac{\sqrt 3 -1}{Y} + \frac{\sqrt 3 +1}{X}=4 \sqrt 2$$ with the unit circle $$X^2+Y^2 = 1.$$

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  • $\begingroup$ How did you get the second and third equation from the first equation in your answer. I knew that the general solution of $\sin x=\sin a$ is $x=n\pi+(-1)^na$. $\endgroup$ – MrAP Mar 10 '19 at 12:17
  • $\begingroup$ @MrAP. Consider that two supplementary angles have the same sine. So $$\sin(\pi- \alpha) = \sin \alpha.$$ $\endgroup$ – dfnu Mar 10 '19 at 12:19
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    $\begingroup$ Oh. I got it. Mine generalizes both odd and even n in one equation. $\endgroup$ – MrAP Mar 10 '19 at 12:20
  • $\begingroup$ @MrAP. Yep. I just prefer to use basic definitions instead of formulae. More ductile, I believe. $\endgroup$ – dfnu Mar 10 '19 at 12:21

Use $$\sin15^{\circ}=\frac{\sqrt3-1}{2\sqrt2}$$ and $$\cos15^{\circ}=\frac{\sqrt3+1}{2\sqrt2}.$$ We obtain: $$\sin(15^{\circ}+x)=\sin2x.$$

Thus, $$15^{\circ}+x=2x+360^{\circ}k,$$ where $k$ is an integer number, or $$15^{\circ}+x=180^{\circ}-2x+360^{\circ}k.$$ Can you end it now?

  • $\begingroup$ I had got that equation at the end. Then I equated the angles to get $x=15^\circ$ $\endgroup$ – MrAP Mar 10 '19 at 12:02
  • $\begingroup$ @MrAP I got also $55^{\circ}.$ $\endgroup$ – Michael Rozenberg Mar 10 '19 at 12:04
  • $\begingroup$ How did you get that value? $\endgroup$ – MrAP Mar 10 '19 at 12:07
  • $\begingroup$ @MrAP I added something. See now. $\endgroup$ – Michael Rozenberg Mar 10 '19 at 12:07

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