Question

Sin x + sin 3x + sin 5x = 0

Answer 1

$\displaystyle sin x + sin 3x + sin 5x = 0 \\ \\ sin3x+(sin \ 5x+sin \ x)=0 \\ \\ sin \ 3x+ 2 sin (\frac{5x+x}{2} ) \ cos \ (\frac{5x-x}{2} ) \\ \\ sin \ 3x+2sin \frac{6x}{2} \cdot cos \frac{4x}{2} \\ \\ sin \ 3x+ 2 \sin \ \ 3cos2x \\ \\ sin \ 3x(1+2cos2x)=0 \\ \\ sin \ 3x=0=\ \textgreater \ 3x=n \pi \\ \\ \boxed{x= \frac{n \pi}{3} }$

Answer 2

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)