# Answer in Trigonometry for aisme #251971

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

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

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

“sin(3x) + sin(x)=2sin (frac{3x+x}2)cos (frac{3x-x}2)=2sin (2x)cos (x)\=2cdot2sin(x)cos(x)cos(x)=4sin(x)cos^2(x).”

(b) Let us find all the angles between 0 and “pi” which satisfy the equation “sin(3x) + sin(x) = 2cos^2(x),” which is equivalent to “4sin(x)cos^2(x)=2cos^2(x),” and hence to “2cos^2(x)(2sin(x)-1)=0.” It follows that “cos(x)=0” or “sin(x)=frac{1}2,” and hence “x=frac{pi}2+pi n” or “x=frac{pi}6+2pi n” or “x=frac{5pi}6+2pi n, ninZ.” Therefore, the angles between 0 and “pi” which satisfy the equation are “frac{pi}2, frac{pi}6, frac{5pi}6.”