1]Show that 2[sinxcosx - 1/2 ] = sin2x -1 2] Hence, determine the general solution for: 2[sinxcosx - 1/2 ] =0
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$\displaystyle 2(sinxcosx-\frac{1}{2}) = 2sinxcosx-2*\frac{1}{2} = 2sinxcosx-1 $ We know that $\displaystyle sin2x=2sinxcosx$ so we have $\displaystyle 2sinxcosx-1 = sin2x-1$
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