Difference between revisions of "1962 IMO Problems/Problem 4"
Brut3Forc3 (talk | contribs) m (Prettified the cosines (\cos, not cos) |
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==Solution== | ==Solution== | ||
− | {{ | + | First, note that we can write the left hand side as a cubic function of <math>\cos^2 x</math>. So there are at most <math>3</math> distinct values of <math>\cos^2 x</math> that satisfy this equation. Therefore, if we find three values of <math>x</math> that satisfy the equation and produce three different <math>\cos^2 x</math>, then we found all solutions to this cubic equation (without expanding it, which is another viable option). Indeed, we find that <math>\frac{\pi}2</math>, <math>\frac{\pi}4</math>, and <math>\frac{\pi}6</math> all satisfy the equation, and produce three different values of <math>\cos^2 x</math>, namely <math>0</math>, <math>\frac12</math>, and <math>\frac34</math>. So we solve <math>\cos^2 x = \text{each of these}</math>. Therefore, our solutions are: |
+ | |||
+ | <math>x = \frac{(2k+1)\pi}2,\, \frac{(2k+1)\pi}4,\, \frac{(6k+1)\pi}6,\, \frac{(6k+5)\pi}6 \quad \forall k\in Z</math> | ||
==See Also== | ==See Also== | ||
{{IMO box|year=1962|num-b=3|num-a=5}} | {{IMO box|year=1962|num-b=3|num-a=5}} |
Latest revision as of 23:45, 18 February 2009
Problem
Solve the equation .
Solution
First, note that we can write the left hand side as a cubic function of . So there are at most distinct values of that satisfy this equation. Therefore, if we find three values of that satisfy the equation and produce three different , then we found all solutions to this cubic equation (without expanding it, which is another viable option). Indeed, we find that , , and all satisfy the equation, and produce three different values of , namely , , and . So we solve . Therefore, our solutions are:
See Also
1962 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |