Question

sin3x/sinx-sinx/sin3x=2cos2x

Answer 1

$frac{sin3x}{sinx} - frac{sinx}{sin3x}=2cos2x \ \ frac{sin^23x-sin^2x}{sinx*sin3x}=2cos2x \ \ frac{(sin3x-sinx)(sin3x+sinx)}{sinx*sin3x}=2cos2x\ \ frac{2sinx*cos2x*2sin2x*cosx}{sinx*sin3x}=2cos2x\ \ left { {{frac{2cos2x*sin2x*cosx}{sin3x}=cos2x} atop {sinx neq 0}} right. \ \ left { {{frac{cos2x*(2sin2x*cosx-sin3x)}{sin3x}=0} atop {sinx neq 0}} right. \ \ left { {{frac{cos2x*(2sin2x*cosx-sin(2x+x))}{sin3x}=0} atop {sinx neq 0}} right.$
$left { {{frac{cos2x*(2sin2x*cosx-(sin2x*cosx+cos2x*sinx))}{sin3x}=0} atop {sinx neq 0}} right. \ \ left { {{frac{cos2x*(sin2x*cosx-cos2x*sinx)}{sin3x}=0} atop {sinx neq 0}} right. \ \ left { {frac{cos2x*sin(2x-x)}{sin3x}=0} }atop {sinx neq 0}} right. \ \ left { {{frac{cos2x*sinx}{sin3x}=0} atop {sinx neq 0}} right. \ \ { {{frac{(1-2sin^2x)*sinx}{sin3x}=0} atop {sinx neq 0}} right.$
${ {{sin^2x= frac{1}{2}} atop {sinx neq 0,sin3x neq 0}} right. \ \ { {{sinx= (+/-)frac{ sqrt{2} }{2}} atop {sinx neq 0,sin3x neq 0}} right. \ \ { {{x= frac{ pi }{4}+ frac{ pi n}{2} ,nEZ} atop {x neq frac{ pi k}{3},kEZ }} right.$
x = π/4 + πn/2, n ∈ Z