Question #251971

(a) Show that sin(3x) + sin(x) = 4sin(x)cos^2(x)

(b) Find all the angles between 0 and π\pi which satisfy the equation sin(3x) + sin(x) = 2(cos^2)(x).


1
Expert's answer
2021-10-18T07:41:21-0400

(a) Let us show that sin(3x)+sin(x)=4sin(x)cos2(x).\sin(3x) + \sin(x) = 4\sin(x)\cos^2(x). Indeed,


sin(3x)+sin(x)=2sin(3x+x2)cos(3xx2)=2sin(2x)cos(x)=22sin(x)cos(x)cos(x)=4sin(x)cos2(x).\sin(3x) + \sin(x)=2\sin (\frac{3x+x}2)\cos (\frac{3x-x}2)=2\sin (2x)\cos (x)\\=2\cdot2\sin(x)\cos(x)\cos(x)=4\sin(x)\cos^2(x).


(b) Let us find all the angles between 0 and π\pi which satisfy the equation sin(3x)+sin(x)=2cos2(x),\sin(3x) + \sin(x) = 2\cos^2(x), which is equivalent to 4sin(x)cos2(x)=2cos2(x),4\sin(x)\cos^2(x)=2\cos^2(x), and hence to 2cos2(x)(2sin(x)1)=0.2\cos^2(x)(2\sin(x)-1)=0. It follows that cos(x)=0\cos(x)=0 or sin(x)=12,\sin(x)=\frac{1}2, and hence x=π2+πnx=\frac{\pi}2+\pi n or x=π6+2πnx=\frac{\pi}6+2\pi n or x=5π6+2πn, nZ.x=\frac{5\pi}6+2\pi n, \ n\in\Z. Therefore, the angles between 0 and π\pi which satisfy the equation are π2, π6, 5π6.\frac{\pi}2,\ \frac{\pi}6,\ \frac{5\pi}6.


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