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

Solve the equation in real number.

2+2+2+x+322+2+x=2x\sqrt{2+\sqrt{2+\sqrt{2+x}}} + \sqrt{3}\sqrt{2-\sqrt{2+\sqrt{2+x}}} = 2x

I've got an idea that 0x20 \leq x \leq 2, but I can't continue further.

Help me!

Note by Samuraiwarm Tsunayoshi
4 years, 4 months ago

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Substitute x=2cos(θ)x=2\cos(\theta)

\therefore 2+2+2+2cos(θ)+322+2+2cos(θ)=4cos(θ)\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+2cos(θ)=2(1+cos(θ))=4cos2(θ/2)2+2\cos(\theta)= 2(1+\cos(\theta))= 4cos^{2}(\theta/2)

Hence are equation becomes: 2+2+2cos(θ/2)+322+2cos(θ/2)=4cos(θ)\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 1cos(θ/4)=2sin2(θ/8)1-\cos(\theta/4)=2\sin^{2}(\theta/8)

So finally we arrive at : 2cos(θ/8)+23sin(θ/8)=4cos(θ) 2|\cos(\theta/8)|+2\sqrt{3}|\sin(\theta/8)|=4\cos(\theta)

\therefore 12cos(θ/8)+32sin(θ/8)=cos(θ)\dfrac{1}{2}|\cos(\theta/8)|+ \dfrac{\sqrt{3}}{2}|\sin(\theta/8)|=\cos(\theta)

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

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

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

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

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

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

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

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

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

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

Samarpit Swain - 4 years, 4 months ago

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Great work!!

However, you will need to be careful, that cos2θ=cosθ \sqrt{ \cos^2 \theta } = | \cos \theta | . So, after the last step, proceed with caution.

Note that in the second line, the RHS should be 4cos2(θ/2) 4 \cos ^2 ( \theta / 2 ) .

Calvin Lin Staff - 4 years, 4 months ago

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

Samarpit Swain - 4 years, 4 months ago

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Sir, but 0x20\leq x \leq 2, Hence 0y10\leq y \leq 1, where y=cos(θ)y=\cos(\theta) So the modulus here has no importance, right?

Samarpit Swain - 4 years, 4 months ago

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@Samarpit Swain Firstly, I disagree with Samuraiwarm's claim that 0x2 0 \leq x \leq 2 . Looking at the domain of the most nested square root, we get x+20 x +2 \geq 0 . We need to look further at 22+2+x \sqrt{ 2 - \sqrt{ 2 + \sqrt{ 2 + x } } } (requires some work) to be able to conclude that x2 x \leq 2 .

Putting this together, we actually have 2x2 -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 θ=7π4 \theta = \frac{ 7 \pi } { 4} was a solution. Is this truly a solution? The issue arises because even though cosθ>0 \cos \theta > 0 , we actually end up with cosθ2<0 \cos \frac{ \theta } { 2} < 0 . As such, we actually have 2+2cosθ=cosθ2cosθ2 \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 - 4 years, 4 months ago

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@Calvin Lin 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?

Samarpit Swain - 4 years, 4 months ago

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

Samuraiwarm Tsunayoshi - 4 years, 4 months ago

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@Samuraiwarm Tsunayoshi Oh, I missed that. Thanks!

Calvin Lin Staff - 4 years, 4 months ago

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Try some trigonometry.

Sudeep Salgia - 4 years, 4 months ago

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I haven't been able to work on this, Sir, But the range 0x20\le x\le2 made me think of the substitution x=X+1x=X+1. Therefore, X[1,1]X\in[-1,1] which immediately reminds one of sinθ\sin\theta or cosθ\cos\theta.

Ishan Dasgupta Samarendra - 4 years, 4 months 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.

Satvik Choudhary - 4 years, 4 months ago

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

Ishan Dasgupta Samarendra - 4 years, 4 months ago

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Hint: Conjugates.

Calvin Lin Staff - 4 years, 4 months ago

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