Problem:
Suppose a is a real number such that the equation
aβ
(sinx+sin(2x))=sin(3x)
has more than one solution in the interval (0,Ο). The set of all such a that can be written in the form
(p,q)βͺ(q,r),
where p,q, and r are real numbers with p<q<r. What is p+q+r ?
Answer Choices:
A. β4
B. β1
C. 0
D. 1
E. 44
Solution:
Observe that
sin(3x)β=sinxβ
cos(2x)+sin(2x)β
cosx=sinxβ
(cos(2x)+2cos2x)=sinxβ
(4cos2xβ1)β
Thus the given equation is equivalent to aβ
sinxβ
(1+2cosx)=sinxβ
(4cos2xβ1). Because sinxξ =0 for x in the interval (0,Ο), the equation reduces to aβ
(1+2cosx)=4cos2xβ1, which is equivalent to
(aβ(2cosxβ1))(1+2cosx)=0
Thus cosx=β21β or cosx=2a+1β.
Recall that the function y=cosx strictly decreases from 1 to -1 in the interval [0,Ο]. Thus for any tβ(β1,1), the equation cosx=t has exactly one solution in (0,Ο). It follows that the original equation has more than one solution precisely when
β1<2a+1β<1 and 2a+1βξ =β21β.
This means aβ(β3,β2)βͺ(β2,1). The requested sum is β3+(β2)+1=(A)β4β.
The problems on this page are the property of the MAA's American Mathematics Competitions