Find the general solution of the equation sin⁡x+sin⁡2x+sin⁡3x=0. I have started doing this problem by applying the formula of sin⁡A+sin⁡B but couldn't generalise it. Please solve it for me.

Question

Find the general solution of the equation sin⁡x+sin⁡2x+sin⁡3x=0. I have started doing this problem by applying the formula of sin⁡A+sin⁡B but couldn&#039;t generalise it. Please solve it for me.

GaceCoect5v · Accepted Answer

HINT:   Using Prosthaphaeresis Formulas,  sin⁡x+sin⁡3x=2sin⁡3x+x2cos⁡3x−x2  We can also use sin⁡x=sin⁡(2x−−x),  sin⁡3x=sin⁡(2x+x)   Now sin⁡y=0⇒y=nπ   and cos⁡A=cos⁡B⇒A=2mπ±B where m,n are arbitrary integers

vrangett · Accepted Answer

You actually can also do this pretty well by just the simple addition formulas from geometry. By that I mean  sin⁡(x+y)=sin⁡(x)cos⁡(y)+cos⁡(x)sin⁡(y)  cos⁡(x+y)=cos⁡(x)cos⁡(y)−sin⁡(x)sin⁡(y)  If you apply these to the expressions sin⁡(2x) and sin⁡(3x)=sin⁡(2x+x) you get after some calculation  f(x)=sin⁡(x)+sin⁡(2x)+sin⁡(3x)=sin⁡(x)cos⁡(x)(2cos⁡(x)+1)  Since the function f is 2π-periodic it suffices to find the solutions on the interval [0,2π). These are just given by the zeros of each factor, which are  A={0,π2,π3π2,acos⁡(−0.5),−acos⁡(0.5)+2π}  So the general solution set is given by A+2πZ

nick1337 · Accepted Answer

(sin⁡x+sin⁡3x)+sin⁡2x=02sin⁡2xcos⁡x+sin⁡2x=0sin⁡2x(2cos⁡x+1)=0sin⁡2x=0 or cos⁡x=−122x=nπ  or  x=nπ2  or  x=2nπ+−2π3

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

Find the general solution of the equation \sin x+\sin 2x+\sin

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