Math, asked by pushpitbhargava85, 5 months ago

sin2x-sin4x+sin6x=0​

Answers

Answered by KrishnaNaveen
2

Step-by-step explanation:

⇒(sin2x+sin6x)−sin4x=0

⇒2sin(

2

2x+6x

)cos(

2

2x−6x

)−sin4x=0

⇒2sin4xcos2x−sin4x=0

⇒sin4x(2cos2x−1)=0

⇒sin4x=0,2cos2x−1=0

⇒sin4x=0,cos2x=

2

1

∴4x=nπ,2x=2nπ±

3

π

∴x=

4

,nπ±

6

π

where n∈Z

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