1.

Solve `sin 2x - sin 4x + sin 6x = 0`.

Answer» `sin2x- sin4x+ sin6x = 0`
as we know that, `(sina + sinb = 2sin((a+b)/2)cos((a-b)/2))`
so, `2sin4x.cos(-2x) - sin4x = 0`
`= Sin4x(2cos2x-1) = 0`
now it can possible that whether `sin4x = 0 or 2cos2x-1=0`
when `sin 4x=0`
`4x = n pi & n in z`
`x= npi/4`
when `2cos 2x-1=0`
`cos2x= cos (pi/3) `
`2x = 2n pi +- pi/3`
`x= n pi +- pi/6`
answer = `(npi)/4 , npi +- pi/6`


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