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!

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!

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TopNewestSubstitute \(x=2\cos(\theta)\)

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

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. – Samarpit Swain · 2 years, 1 month ago

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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 \( 4 \cos ^2 ( \theta / 2 ) \). – Calvin Lin Staff · 2 years, 1 month ago

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– Samarpit Swain · 2 years, 1 month 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?Log in to reply

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. – Calvin Lin Staff · 2 years, 1 month ago

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– Samuraiwarm Tsunayoshi · 2 years, 1 month ago

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

– Calvin Lin Staff · 2 years, 1 month ago

Oh, I missed that. Thanks!Log in to reply

– Samarpit Swain · 2 years, 1 month 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?Log in to reply

– Samarpit Swain · 2 years, 1 month ago

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

Try some trigonometry. – Sudeep Salgia · 2 years, 1 month ago

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– Ishan Dasgupta Samarendra · 2 years, 1 month 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 \(\cos\theta\).Log in to reply

– Satvik Choudhary · 2 years, 1 month ago

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

– Ishan Dasgupta Samarendra · 2 years, 1 month ago

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

Hint:Conjugates. – Calvin Lin Staff · 2 years, 1 month agoLog in to reply