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# 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
2 years, 3 months ago

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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. · 2 years, 3 months ago

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 · 2 years, 3 months ago

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? · 2 years, 3 months ago

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 · 2 years, 3 months ago

Consider LHS; both sides are square roots, which is always non-negative. That means RHS is always non-negative as well. · 2 years, 3 months ago

Oh, I missed that. Thanks! Staff · 2 years, 3 months ago

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? · 2 years, 3 months ago

Oh yes, I had completely forgotten that, Thanks a ton sir, for your inputs! · 2 years, 3 months ago

Try some trigonometry. · 2 years, 3 months ago

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 $$\cos\theta$$. · 2 years, 3 months ago

I think it would be better if you would cosider x=2X rather than x=X+1 and then you can use trigo. · 2 years, 3 months ago

Perhaps. Like I said, I haven't tried this at all and won't be able to till around late evening. · 2 years, 3 months ago

Hint: Conjugates. Staff · 2 years, 3 months ago

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