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olganol [36]
3 years ago
12

Sinx+sin3x=0 sin5x=-cos2x

Mathematics
1 answer:
Rasek [7]3 years ago
8 0

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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