Number of solutions

Answer :-(

I 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]

We 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

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.