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If sin 2x + sin x = 0, then x is (A)npi...

If sin 2x + sin x = 0, then x is (A)`npi + pi/3, 2npi + pi/6` (B) `npi , npi + (-1)^n pi/3` (C) `npi, 2npi + 2pi/3, 2npi - 2pi/3` (D) `2npi, npi + 2pi/3`

Text Solution

Verified by Experts

`sin2x+sinx = 0`
`=>2sinxcosx+sinx = 0`
`=>sinx(2cosx+1) = 0`
So, `sinx = 0` and `2cosx+1 = 0`
`=>sinx = 0` and `cosx = -1/2`
`=> x = npi ` and `x = 2npi+-(2pi)/3`
So, option `C` is the correct option.
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