Problem B. 4138. (December 2008)

B. 4138. Solve the following equation: $\sqrt{2}\cdot (\sin x+\cos x)= \tan x+ \cot x$ .

(4 pont)

Deadline expired on January 15, 2009.

Sorry, the solution is available only in Hungarian. Google translation

Megoldás: Ha a jobboldal értelmes, akkor $\mathop{\rm ctg} x=1/\mathop{\rm tg} x$ miatt $|\mathop{\rm tg} x+ \mathop{\rm ctg} x|\ge 2$ . Másrészről

$\sin x+\cos x=2\cos(45^\circ-x)\sin45^\circ=\sqrt{2}\cos(45^\circ-x)$

miatt $|\sqrt{2}\cdot (\sin x+\cos x)|=2|\cos(45^\circ-x)|\le 2$ . Ezért az egyenlet minden x megoldására |cos (45^o-x)|=1, vagyis x=45^o+k^.180^o teljesül alkalmas k egész számmal. Ezek közül azonban csak a 45^o+k^.360^o alakú szögek megoldásai az egyenletnek.

Statistics:

132 students sent a solution.

4 points: 76 students.

3 points: 37 students.

2 points: 8 students.

1 point: 2 students.

0 point: 3 students.

Unfair, not evaluated: 6 solutionss.

Problems in Mathematics of KöMaL, December 2008

132 students sent a solution.
4 points:	76 students.
3 points:	37 students.
2 points:	8 students.
1 point:	2 students.
0 point:	3 students.
Unfair, not evaluated:	6 solutionss.

Already signed up?
New to KöMaL?