(a) Let us show that sin(3x)+sin(x)=4sin(x)cos2(x). Indeed,
sin(3x)+sin(x)=2sin(23x+x)cos(23x−x)=2sin(2x)cos(x)=2⋅2sin(x)cos(x)cos(x)=4sin(x)cos2(x).
(b) Let us find all the angles between 0 and π which satisfy the equation sin(3x)+sin(x)=2cos2(x), which is equivalent to 4sin(x)cos2(x)=2cos2(x), and hence to 2cos2(x)(2sin(x)−1)=0. It follows that cos(x)=0 or sin(x)=21, and hence x=2π+πn or x=6π+2πn or x=65π+2πn, n∈Z. Therefore, the angles between 0 and π which satisfy the equation are 2π, 6π, 65π.
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