# Any ideas on solving this equation? (help meh!)

Solve the equation in real number.

$\sqrt{2+\sqrt{2+\sqrt{2+x}}} + \sqrt{3}\sqrt{2-\sqrt{2+\sqrt{2+x}}} = 2x$

I've got an idea that $0 \leq x \leq 2$, but I can't continue further.

Help me! Note by Samuraiwarm Tsunayoshi
6 years ago

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## Comments

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Substitute $x=2\cos(\theta)$

$\therefore$ $\sqrt{2+\sqrt{2+\sqrt{2+2\cos(\theta)}}} + \sqrt{3}\sqrt{2-\sqrt{2+\sqrt{2+2\cos(\theta)}}}= 4\cos(\theta)$

Now, $2+2\cos(\theta)= 2(1+\cos(\theta))= 4cos^{2}(\theta/2)$

Hence are equation becomes: $\sqrt{2+\sqrt{2+2|\cos(\theta/2)|}}+ \sqrt{3}\sqrt{2-\sqrt{2+2|\cos(\theta/2)|}}=4\cos(\theta)$

Repeat this process till you eliminate the radicals( it will take you two more steps).

Note: In the last step you will have to use $1-\cos(\theta/4)=2\sin^{2}(\theta/8)$

So finally we arrive at : $2|\cos(\theta/8)|+2\sqrt{3}|\sin(\theta/8)|=4\cos(\theta)$

$\therefore$ $\dfrac{1}{2}|\cos(\theta/8)|+ \dfrac{\sqrt{3}}{2}|\sin(\theta/8)|=\cos(\theta)$

$\text{CASE 1: When \theta lies in the first quadrant}$

$\bullet$ $\cos(\theta/8-\pi/3)=cos(\theta)$

$\text{CASE 2: When theta lies in the second quadrant}$

$\bullet$ $-\cos(\theta/8+\pi/3)= cos(\theta)$

$\text{CASE 3: When theta lies in the third quadrant}$

$\bullet$ $-\cos(\theta/8-\pi/3)=cos(\theta)$

$\text{CASE 4: When theta lies in the fourth quadrant}$

$\bullet$ $\cos(\theta/8+\pi/3)= cos(\theta)$

Solve this general equation for $\theta$, take the union of all cases and plug the permissible values in $x=2\cos(\theta)$, to find your desired solutions.

Feel free to ask if you have a doubt at any point.

- 6 years ago

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Great work!!

However, you will need to be careful, that $\sqrt{ \cos^2 \theta } = | \cos \theta |$. So, after the last step, proceed with caution.

Note that in the second line, the RHS should be $4 \cos ^2 ( \theta / 2 )$.

Staff - 6 years ago

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Oh yes, I had completely forgotten that, Thanks a ton sir, for your inputs!

- 6 years ago

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Sir, but $0\leq x \leq 2$, Hence $0\leq y \leq 1$, where $y=\cos(\theta)$ So the modulus here has no importance, right?

- 6 years ago

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Firstly, I disagree with Samuraiwarm's claim that $0 \leq x \leq 2$. Looking at the domain of the most nested square root, we get $x +2 \geq 0$. We need to look further at $\sqrt{ 2 - \sqrt{ 2 + \sqrt{ 2 + x } } }$ (requires some work) to be able to conclude that $x \leq 2$.

Putting this together, we actually have $-2 \leq x \leq 2$. So, you should verify the assumptions, especially when it's phrased as "I have an idea that this might work". It just might be possible that we have negative solutions.

Secondly, (for sake of argument) suppose you found that $\theta = \frac{ 7 \pi } { 4}$ was a solution. Is this truly a solution? The issue arises because even though $\cos \theta > 0$, we actually end up with $\cos \frac{ \theta } { 2} < 0$. As such, we actually have $\sqrt{ 2 + 2 \cos \theta } = - \cos \frac{ \theta} {2} \neq \cos \frac{ \theta}{2}$ IE we need to take the negative square root.

Staff - 6 years ago

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Actually i thought it was given as a constraint along with the equation. Sir i've made the necessary changes. Can you check it once, please?

- 6 years ago

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Consider LHS; both sides are square roots, which is always non-negative. That means RHS is always non-negative as well.

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Oh, I missed that. Thanks!

Staff - 6 years ago

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Try some trigonometry.

- 6 years ago

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I haven't been able to work on this, Sir, But the range $0\le x\le2$ made me think of the substitution $x=X+1$. Therefore, $X\in[-1,1]$ which immediately reminds one of $\sin\theta$ or $\cos\theta$.

- 6 years ago

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I think it would be better if you would cosider x=2X rather than x=X+1 and then you can use trigo.

- 6 years ago

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Perhaps. Like I said, I haven't tried this at all and won't be able to till around late evening.

- 6 years ago

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Hint: Conjugates.

Staff - 6 years ago

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