Sin x + sin 3x + sin 5x = 0
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Answer:
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)
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