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QUESTION
i dont get how you solve this . 4sinxcosx = square root of (3) ; x = [0,2pi)?
i dont get how you solve this .4sinxcosx = square root of (3) ; x = [0,2pi)?4sinxcosx = √32*2sinxcosx = √32*sin2x = √3sin2x = √3 / 2sin2x = sin(π / 3), sin(2π / 3)=>2x = π / 3, 2π / 3x = π / 6, 2π / 6..
