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Difference between revisions of "1965 IMO Problems/Problem 1"
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Case 2: <math>\cos 2x</math> is negative. This means that <math>x</math> is in one of the interavals <math>(\pi/4, \pi/2)</math> or <math>(3\pi/2, 7\pi/4)</math>. We must find all <math>x</math> in these two intervals such that <math>2\cos^2 x\leq 1+\cos 2x</math>, which is equivalent to <math>2\cos^2 x\leq 2\cos^2 x</math>, which is true for all <math>x</math> in these intervals.
 
Case 2: <math>\cos 2x</math> is negative. This means that <math>x</math> is in one of the interavals <math>(\pi/4, \pi/2)</math> or <math>(3\pi/2, 7\pi/4)</math>. We must find all <math>x</math> in these two intervals such that <math>2\cos^2 x\leq 1+\cos 2x</math>, which is equivalent to <math>2\cos^2 x\leq 2\cos^2 x</math>, which is true for all <math>x</math> in these intervals.
  
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Therefore the left inequality is true when <math>x</math> is in one of the intervals <math>[\pi/4, \pi/2)</math> or <math>(3\pi/2, 7\pi/4]</math>. We shall now deal with the right inequality.
+
Therefore the left inequality is true when <math>x</math> is in the union of the intervals <math>[\pi/4, \pi/2)</math>, <math>(3\pi/2, 7\pi/4]</math>, and <math>[\pi/2, 3\pi/2]</math>, which is the interval <math>[\pi/4, 7\pi/4]</math>. We shall now deal with the right inequality.
  
−
As above, we can square it and have it be true whenever the original right inequality is true, so we do that. <math>2-2\sqrt{\cos^2{2x}}\leq 2</math>, which is always true. Therefore the original right inequality is always satisfied, and all <math>x</math> such that the original inequality is satisfied are in the intervals <math>[\pi/4, \pi/2)</math> and <math>(3\pi/2, 7\pi/4]</math>.
+
As above, we can square it and have it be true whenever the original right inequality is true, so we do that. <math>2-2\sqrt{\cos^2{2x}}\leq 2</math>, which is always true. Therefore the original right inequality is always satisfied, and all <math>x</math> such that the original inequality is satisfied are in the interval <math>[\pi/4, 7\pi/4)</math>.

Revision as of 10:34, 9 July 2010

Problem

Determine all values $x$ in the interval $0\leq x\leq 2\pi$ which satisfy the inequality \[2\cos x \leq \left| \sqrt{1+\sin 2x} - \sqrt{1-\sin 2x } \right| \leq \sqrt{2}.\]

Solution

We shall deal with the left side of the inequality first ($2\cos x \leq \left| \sqrt{1+\sin 2x} - \sqrt{1-\sin 2x } \right|$) and the right side after that.

It is clear that the left inequality is true when $\cos x$ is non-positive, and that is when $x$ is in the interval $[\pi/2, 3\pi/2]$. We shall now consider when $\cos x$ is positive. We can square the given inequality, and the resulting inequality will be true whenever the original left inequality is true. $4\cos^2{x}\leq 1+\sin 2x+1-\sin 2x-2\sqrt{1-\sin^2 2x}=2-2\sqrt{\cos^2{2x}}$. This inequality is equivalent to $2\cos^2 x\leq 1-\left| \cos 2x\right|$. I shall now divide this problem into cases.

Case 1: $\cos 2x$ is non-negative. This means that $x$ is in one of the intervals $[0,\pi/4]$ or $[7\pi/4, 2\pi]$. We must find all $x$ in these two intervals such that $2\cos^2 x\leq 1-\cos 2x$. This inequality is equivalent to $2\cos^2 x\leq 2\sin^2 x$, which is only true when $x=\pi/4$ or $7\pi/4$.

Case 2: $\cos 2x$ is negative. This means that $x$ is in one of the interavals $(\pi/4, \pi/2)$ or $(3\pi/2, 7\pi/4)$. We must find all $x$ in these two intervals such that $2\cos^2 x\leq 1+\cos 2x$, which is equivalent to $2\cos^2 x\leq 2\cos^2 x$, which is true for all $x$ in these intervals.

Therefore the left inequality is true when $x$ is in the union of the intervals $[\pi/4, \pi/2)$, $(3\pi/2, 7\pi/4]$, and $[\pi/2, 3\pi/2]$, which is the interval $[\pi/4, 7\pi/4]$. We shall now deal with the right inequality.

As above, we can square it and have it be true whenever the original right inequality is true, so we do that. $2-2\sqrt{\cos^2{2x}}\leq 2$, which is always true. Therefore the original right inequality is always satisfied, and all $x$ such that the original inequality is satisfied are in the interval $[\pi/4, 7\pi/4)$.

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