Problem:
How many solutions does the equation sin(2Οβcosx)=cos(2Οβsinx) have in the closed interval [0,Ο] ?
Answer Choices:
A. 0
B. 1
C. 2
D. 3
E. 4
Solution:
Let x be a solution to the given equation in the interval [0,Ο]. Then 2Οβsinxβ[0,2Οβ], so cos(2Οβsinx)β₯0, and therefore by the given equation, sin(2Οβcosx)β₯0. Likewise, xβ[0,Ο] implies that 2Οβcosxβ[β2Οβ,2Οβ], but because sin(2Οβcosx)β₯0, it follows that 2Οβcosx is also in [0,2Οβ].
Because cosΞ±=sin(2ΟββΞ±), the given equation is equivalent to
sin(2Οβcosx)=sin(2Οββ2Οβsinx),
so cosx=1βsinx. Squaring both sides of this last equation gives cos2x=1β2sinx+sin2x, so 2sin2xβ 2sinx=2(sinxβ1)sinx=0. From this it follows that x=0 or x=2Οβ or x=Ο. The first two of these values satisfy the original equation, but the third does not. (Squaring introduced this extraneous solution.) Therefore there are exactly 2β solutions in the interval [0,Ο].
The problems on this page are the property of the MAA's American Mathematics Competitions
