No problem, we see these in the 1st and 2nd grade of high school, high school here is 4 years.

OOOOO MY GAD!!! THANK YOU LEO BRO 🥲😁

sin 3x + sin 2x - sin x = 4sin x * ((cos x)/2)
sin 3x = 3sin x - 4sin^3
sin 2x = 2sin x * cos
3sin x - 4sin^3 x + 2sin x * cos x - sin x =
(3sin x - 4sin^3 x + 2sin x * cos x - sin x)
We can simplify the equation with sin x, but we must make sure that we do not miss the case sin x = 0. When we simplify with sin x in the case sin x ≠ 0, we get:
2 + 2cos x - 4sin^2 x = 2cos x * (cos 3x)/2

Let's expand the right side:

2cos x * (cos 3x)/2 = cos x * cos 3x

Using cosine expansion, cos 3x:

cos 3x = 4cos^3 x - 3cos x

If we substitute this expression:
2 + 2cos x - 4sin^2 x =cos x(4 cos^ 3
It may be necessary to examine these expressions in more detail later, but the important thing is to examine the equation under a corner case. In the first case, when sin x = 0, we can solve sinx = 0 ⇒ x = n.

Solution:

↓

x = ηπ, NEZ

I think this is x

Sin3x+sin2x-sinx=4sinx.(cosx÷2)(cos3x÷2)
Trigonometry ~ 😭

Hehe