Question

find the general solution of the equation sin3x+sinx=sin2x

Answer 1

sinx + sin3x + sin5x = 0
sin3x + (sinx + sin5x) = 0
we know,
sinC + sinD = 2sin(C + D)/2.cos(C-D)/2

sin3x + {2sin(x + 5x)/2.cos(5x-x)/2}= 0
sin3x + {2sin3x.cos2x} = 0
sin3x(1+2cos2x) = 0

Here, sin3x = 0 or cos2x = -1/2

sin3x = 0
3x =nπ
x = nπ/3

again,
cos2x = -1/2
cos2x = -cos(π/3)
cos2x = cos(π -π/3)
cos2x = cos(2π/3)

we know,
If cos∅ = cosA then,
∅ = 2nπ ± A

2x = 2nπ ± (2π/3)
x = nπ ± (π/3)

Hence, the solutions are
x = nπ/3 or nπ ±(π/3)

Answer 2

⏏HOPE THIS ⬆ HELPS YOU⏏