# Sum of the squares of the solutions to $\sqrt{1-\cos{x}}+\sqrt{1+\cos{x}} = \sqrt{3}.$

Problem: Find the sum of the squares of the solutions to $\sqrt{1-\cos{x}}+\sqrt{1+\cos{x}} = \sqrt{3},$ where $-\pi<x<\pi.$

Attempt: Squaring both sides gives $$1-\cos{x}+2\sqrt{(1-\cos{x})(1+\cos{x})}+1+\cos{x}=3,$$

which simplifies to $$2\sqrt{1-\cos^2{x}}=3-1\Longleftrightarrow \sqrt{\sin^2{x}} = 1 \Longleftrightarrow |\sin{x}|=1.$$

Questions:

1) Is my procedure so far correct?

2) How do I proceed?

3) Is there any easier way of solving this problem?

• It's $3-2$ and not $3-1$ – kingW3 May 8 '17 at 7:56
• Yes, I saw that. But the problem is the equation $|\sin{x}|=1/2.$ – Parseval May 8 '17 at 7:57
• In general you're solving $\sin x=1/2$ and $\sin x=-1/2$ – kingW3 May 8 '17 at 8:00

Well, when you square the LHS:

$$\left(\sqrt{1-\cos\left(x\right)}+\sqrt{1+\cos\left(x\right)}\right)^2=2+2\cdot\sqrt{1-\cos\left(x\right)}\cdot\sqrt{1+\cos\left(x\right)}=$$ $$2+2\cdot\sqrt{\sin^2\left(x\right)}=2\cdot\left(1+\sqrt{\sin^2\left(x\right)}\right)\tag1$$

So, we get:

$$\left(\sqrt{1-\cos\left(x\right)}+\sqrt{1+\cos\left(x\right)}\right)^2=\sqrt{3}\space\Longleftrightarrow\space2\cdot\left(1+\sqrt{\sin^2\left(x\right)}\right)=3\space\Longleftrightarrow\space$$ $$\sqrt{\sin^2\left(x\right)}=\frac{3}{2}-1=\frac{1}{2}=\left|\sin\left(x\right)\right|\tag2$$

• But how do I solve $|\sin{x}| = 1/2$ in the given interval? I don't know how to treat the absolute value when it contains a sinefunction like that. Should I splitt it up in two cases, just like the definition of the absolute value tells me to? – Parseval May 8 '17 at 7:55
• Yes, split into two cases, plus or minus a half. – Arby May 8 '17 at 8:03
• The interval in which x lies, it is the same as writing $0<x<2\pi$ right? – Parseval May 8 '17 at 8:20

From where you left off...$x = \pm \dfrac{\pi}{2}$, can you finish it ?Oops, you made a mistake, and it should read $|\sin x| = 0.5 \implies x = \pm \dfrac{\pi}{6}, \pm \dfrac{5\pi}{6}$.