1I am sure there is an easier way of doing it, because it is the number of solutions and not solutions itself, but this is what I got in my first attempt.....
sinx+2sin2x=3+sin3x
=> sinx-sin3x=3-2sin2x
=>-2sinx.cos2x=3-4sinx.cosx
=>2sinx(2cosx-cos2x)=3
=>2sinx(2cosx-cos2x+sin2x) =3
=>2t(t2+t2-1+2√1-t2) =3 t=sinx; -1≤t≤1
=>4t3+4t√1-t2-2t-3=0
Solve this cubic eqn in t for values of sinx, and then x
[339]
341We first note that the equation as written implies that LHS≤3. So we must have sin 3x ≤ 0
This means x \in \left[\frac{\pi}{3}, \frac{2 \pi}{3} \right]
Again, we note that since RHS ≥ 2, we must have sin 2x ≥ 1/2 so that we must have
x \in \left[\frac{\pi}{12}, \frac{5 \pi}{12} \right]
The intersection of the above two gives us that a solution if it exists would belong to \left[\frac{\pi}{3}, \frac{5 \pi}{12} \right]
Now
sinx \sinx - \sin 3x + 2 \sin 2x = 4 \sin x \cos x - 2 \sinx \cos 2x = 2 \sin x (2 \cos x - \cos 2x)
2 \cos x - \cos 2x = 2 \cos x (1-\cos x) + 1 \le 2 \times \frac{1}{2} \times \frac{1}{2} + 1 = \frac{3}{2}
[Remember that x(1-x) ≤ 1/4 when 0≤x≤1]
Since in the given interval sin x <1, LHS = 2 \sin x(2 \cos x - \cos 2x) < 2 \times \frac{3}{2} = 3
whereas RHS = 3. Hence, the equation has no solutions
1using graphs wud help...we'd get the ans in a jiffy!!
341
Hari Shankar
·2009-08-26 00:03:26
@shreyan (or anyone who knows) - how do you use graphs for a solution w/o actually using some software. I mean specifically, how would you use graphs in an exam situation for this problem
62prophet sir, you missed a very simple solution
I saw this after seeing your own solution ;)
sin x-\sinx - \sin 3x + 2 \sin 2x = 4 \sin x \cos x - 2 \sinx \cos 2x
= 2(sin 2x-cos 2x) = 2√2 sin(2x-pi/4) = 3
thus, sin(2x-pi/4) = 3/(2√2) > 1
hence no solution :)
341
Hari Shankar
·2009-08-26 07:58:57
you started off from a typo in my solution :D. That bound of 3 is pretty tight.
sinx - sin 3x + 2 sin 2x = 4 sinx cosx - 2 cos 2x sin x
62
Lokesh Verma
·2009-08-27 06:41:08
Prophet sir, Have I made some mistake?
If I have then I am not getting it..
otherwise I am trying to read your joke toooooo seriously [6]
341
Hari Shankar
·2009-08-27 07:35:35
i made the first mistake by forgetting to type in a sin x next to the cos 2x term [you can see I have factorised it next as 2 sin x (2 cos x - cos 2x)]
I was saying that your solution headed off from that point onwards.
62
Lokesh Verma
·2009-08-27 07:38:41
oh yeah.. now i realize [1]
