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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` |
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Answer» `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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