# How to find the roots of $\frac{\sqrt{3}-1}{\sin x}+\frac{\sqrt{3}+1}{\cos x}=4\sqrt{2}$? [duplicate]

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, $$x=\pi/12$$

• How did you find the one value that you found? (Perhaps show your working -- it's possible that you accidentally lost some solutions.) – Minus One-Twelfth Mar 10 '19 at 11:53
• 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 $x=\pi/12$ – MrAP Mar 10 '19 at 11:56
• 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 $\pi/2$. – Minus One-Twelfth Mar 10 '19 at 11:58
• @MinusOne-Twelfth, Check my question now. – MrAP Mar 10 '19 at 12:06
• 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. – 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 $$X^2+Y^2 = 1.$$

• 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 $x=n\pi+(-1)^na$. – MrAP Mar 10 '19 at 12:17
• @MrAP. Consider that two supplementary angles have the same sine. So $$\sin(\pi- \alpha) = \sin \alpha.$$ – dfnu Mar 10 '19 at 12:19
• Oh. I got it. Mine generalizes both odd and even n in one equation. – MrAP Mar 10 '19 at 12:20
• @MrAP. Yep. I just prefer to use basic definitions instead of formulae. More ductile, I believe. – 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?

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