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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## Comments

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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 · 1 year, 11 months 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 · 1 year, 11 months ago

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– Samarpit Swain · 1 year, 11 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?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 · 1 year, 11 months ago

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– Samuraiwarm Tsunayoshi · 1 year, 11 months 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 · 1 year, 11 months ago

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

– Samarpit Swain · 1 year, 11 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?Log in to reply

– Samarpit Swain · 1 year, 11 months ago

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

Try some trigonometry. – Sudeep Salgia · 1 year, 11 months ago

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– Ishan Dasgupta Samarendra · 1 year, 11 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 \(\cos\theta\).Log in to reply

– Satvik Choudhary · 1 year, 11 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.Log in to reply

– Ishan Dasgupta Samarendra · 1 year, 11 months 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 · 1 year, 11 months agoLog in to reply