Hey, can anybody provide a solution for this problem? I am totally clueless.

Q.)Find the possible real values of \((x,y)\) satisfying \[\sin { x } +\cos { x } =\frac { { y }^{ 2 }-2y+4 }{ { y }^{ 2 }+2y+4 }. \]

Hey, can anybody provide a solution for this problem? I am totally clueless.

Q.)Find the possible real values of \((x,y)\) satisfying \[\sin { x } +\cos { x } =\frac { { y }^{ 2 }-2y+4 }{ { y }^{ 2 }+2y+4 }. \]

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TopNewestExactly what @Deeparaj Bhat said... \[\sin { x } +\cos { x } =\frac { { y }^{ 2 }-2y+4 }{ { y }^{ 2 }+2y+4 }=k\]

\(k=\dfrac{y^2-2y+4}{y^2+2y+4}\) \[\implies y^2(k-1)-2y(k+1)+4(k-1)=0....(1)\] Since \(y\in R\) the discriminate of quadratic must be non-negative. \[\implies (k+1)^2-4(k-1)^2\ge 0\] \[\implies (3k-1)(k-3)\le 0\implies k\in[\dfrac 13,3]\]

While \(\sin x+\cos x\in[-\sqrt2,\sqrt 2]\)

Hence \(\boxed{k\in[\dfrac 13,\sqrt 2]}\)

Solve \(1\) using quadratic formula:

\[\boxed{y=\dfrac { 1+k\pm \sqrt { \left( k-3 \right) \left( 3k-1 \right) } }{ 1-k } },k\ne 1\]

While solving \(\sin x+\cos x=k\)

\[\sin \left(x+\dfrac{\pi}{4}\right)=\dfrac k{\sqrt 2}\]

\[\implies x=-\dfrac{\pi}{4}+2n\pi\pm\sin^{-1}\left(\dfrac{k}{\sqrt 2}\right)\]

While when \(k=1\), we get \(y=0\) while \(x=0,\pi/2,2\pi,\dfrac{5\pi}2\cdots\) – Rishabh Cool · 1 year, 1 month ago

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\(\sin x +\cos x \in [-\sqrt{2},\sqrt{2}]\)

The range of the right side is \( \left[ \frac{1}{3},3 \right]\)

So, \(\sin x +\cos x ≥\frac{1}{3}\)

Now replacing \(\sin x =\frac{2t}{1+t^2},\quad \cos x =\frac{1-t^2}{1+t^2}\)

Where \(t=\tan x\), we will then get,

\(2t^2-3t-1≤0 \\ t \in \left[\frac{3-\sqrt{17} }{4}, \frac{3+\sqrt{17}}{4} \right] \\ \therefore x \in \left[ n\pi+\tan^{-1}\left( \frac{3-\sqrt{17}}{4}\right) , n\pi+\tan^{-1}\left( \frac{3+\sqrt{17}}{4}\right)\right]\) – Akshat Sharda · 1 year, 1 month ago

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Let sinx+cosx=k

It is simple to show that k>= -sqrt(2)

Now, k(y^2+2y+4)=y^2 -2y +4

y^2 (k-1) +2y(k+1) +4(k-1) = 0

For y to be real, D >= 0

This implies 1/3 >= k >= -sqrt(2)

Thus for all real values of x such that k<=1/3 , we will get two real values of y.

I can't solve it further, any help appreciated.

Sorry for no latex. – Harsh Shrivastava · 1 year, 1 month ago

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@Akshat Sharda the answer is not correct. – Rishi Sharma · 1 year, 1 month ago

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– Akshat Sharda · 1 year, 1 month ago

Can you please elaborate what I've done wrong?Log in to reply

– Rishi Sharma · 1 year, 1 month ago

Even I am confused. First you also need to find the value of y as well. And lastly the correct answer (from a credible source) is \[x=2n\pi +\cos { \frac { \pi }{ 4 } \pm } \cos ^{ -1 }{ \frac { k }{ \sqrt { 2 } } } \\ y=\frac { 1+k\pm \sqrt { \left( k-3 \right) \left( 3k-1 \right) } }{ 1-k } \quad ;\quad k\in \left[ \frac { 1 }{ 3 } ,\sqrt { 2 } \right] -\left\{ 1 \right\} \]Log in to reply

Now, coming to how to solve it. Put the RHS as \(k\). Note that \(k \leq \sqrt{2}\) due to the LHS.

Also, via calculus (or inequalities) you can see that \(\frac{1}{3} \leq k\) (as \(y\) is real).

Now, finding \(x\) in terms of \(k\) is easy.

Also, for \(k \neq 1\) we can solve for \(y\) as a quadratic. If \(k=1\), then \(y=0\).

And we're done. – Deeparaj Bhat · 1 year, 1 month ago

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– Rishi Sharma · 1 year, 1 month ago

Thanks. Got itLog in to reply

@Deeparaj Bhat @Rishabh Cool @Prakhar Bindal please help – Rishi Sharma · 1 year, 1 month ago

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