expeditiupc

2021-12-26

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

GaceCoect5v

HINT:
Using Prosthaphaeresis Formulas,
$\mathrm{sin}x+\mathrm{sin}3x=2\mathrm{sin}\frac{3x+x}{2}\mathrm{cos}\frac{3x-x}{2}$
We can also use
Now $\mathrm{sin}y=0⇒y=n\pi$
and $\mathrm{cos}A=\mathrm{cos}B⇒A=2m\pi ±B$ where m,n are arbitrary integers

vrangett

You actually can also do this pretty well by just the simple addition formulas from geometry. By that I mean
$\mathrm{sin}\left(x+y\right)=\mathrm{sin}\left(x\right)\mathrm{cos}\left(y\right)+\mathrm{cos}\left(x\right)\mathrm{sin}\left(y\right)$
$\mathrm{cos}\left(x+y\right)=\mathrm{cos}\left(x\right)\mathrm{cos}\left(y\right)-\mathrm{sin}\left(x\right)\mathrm{sin}\left(y\right)$
If you apply these to the expressions $\mathrm{sin}\left(2x\right)$ and $\mathrm{sin}\left(3x\right)=\mathrm{sin}\left(2x+x\right)$ you get after some calculation
$f\left(x\right)\phantom{\rule{0.222em}{0ex}}=\mathrm{sin}\left(x\right)+\mathrm{sin}\left(2x\right)+\mathrm{sin}\left(3x\right)=\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)\left(2\mathrm{cos}\left(x\right)+1\right)$
Since the function f is $2\pi$-periodic it suffices to find the solutions on the interval $\left[0,2\pi \right)$. These are just given by the zeros of each factor, which are
$A\phantom{\rule{0.222em}{0ex}}=\left\{0,\frac{\pi }{2},\pi \frac{3\pi }{2},a\mathrm{cos}\left(-0.5\right),-a\mathrm{cos}\left(0.5\right)+2\pi \right\}$
So the general solution set is given by $A+2\pi \mathbb{Z}$

nick1337

Jeffrey Jordon

Answer is given below (on video)

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