Answer:
We have, sinx+sin3x+sin5x=0
∴(sinx+sin5x)+sin3x=0
∴2sin(
2
x+5x
)cos(
5x−x
)+sin3x=0
∴2sin3xcos2x+sin3x=0
∴sin3x(2cos2x+1)=0
Either sin3x=0 or 2cos2x+1=0
i.e. sin3x=0 or cos2x=−
1
Now, cos2x=−cos
3
π
∴cos2x=cos(π−
)
∴cos2x=cos
2π
∴sin3x=0 or cos2x=cos
3x=nπ,n∈Z or 2x=2mπ±
where m∈Z
Hence, x=
nπ
or x=mπ±
, where n,m∈Z.
6 and -12
Step-by-step explanation:
6×-12=-72
6+(-12)=-6
×^2 -6x-72=0
(x+6) (x-12)
x^2-6x-72=(x+6) (X-12)