Find the solution of sin X= -√3÷2
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Answer:
We have, sinx+sin3x+sin5x=0
∴(sinx+sin5x)+sin3x=0
∴2sin(
2
x+5x
)cos(
2
5x−x
)+sin3x=0
∴2sin3xcos2x+sin3x=0
∴sin3x(2cos2x+1)=0
Either sin3x=0 or 2cos2x+1=0
i.e. sin3x=0 or cos2x=−
2
1
Now, cos2x=−cos
3
π
∴cos2x=cos(π−
3
π
)
∴cos2x=cos
3
2π
∴sin3x=0 or cos2x=cos
3
2π
3x=nπ,n∈Z or 2x=2mπ±
3
2π
where m∈Z
Hence, x=
3
nπ
or x=mπ±
3
π
, where n,m∈Z.
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