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If $displaystyle frac1sin frac3pin=frac1sin frac5pin,nin mathbbZ$, then number of $n$ satisfies given equation ,is

What I tried:

Let $displaystyle fracpin=x$ and equation is $sin 5x=sin 3x$

$displaystyle sin (5x)-sin (3x)=2sin (4x)cos (x)=0$

$displaystyle 4x= mpi$ and $displaystyle x= 2mpipm fracpi2$

$displaystyle frac4pin=mpiRightarrow n=frac4min mathbbZ$ for $m=pm 1,pm 2pm 3,pm 4$

put into $displaystyle x=2mpipm fracpi2$

How do i solve my sum in some easy way Help me please

No need to go with sum-to-product formulas. From $sin5x=sin3x$ you get either
$$
5x=3x+2kpi
$$
or
$$
5x=pi-3x+2kpi
$$
The former yields $x=kpi$, so $1=kn$, whence $n=pm1$.

The latter yields
$$
fracpin=frac18(2k+1)pi
$$
that is, $(2k+1)n=8$. Hence $n=pm8$.

I'm afraid you have used the wrong property: $sin A-sin B=2cosleft(dfracA+B2right)sinleft(dfracA-B2right)$. So you get$$2cos(4x)sin x=0$$Thus, $x=kpivee4x=(2k+1)pi/2$ for $kinBbb Z$. For the former, $x=pi/n=kpiimplies nk=1$ which is true iff $n=k=1vee n=k=-1$. In the latter case, $(2k+1)pi/8=pi/nimplies(2k+1)n=8$ which is true iff $n=-8,k=-1vee n=8,k=0$.

Also, remember that $sin 3x$ and $sin 5x$ have to be non-zero, because they are in the denominator in the original problem statement. So we have to reject $pm1$ to get two solutions, $n=pm8$.

Since $-n$ is a solution if $n$ is a solution, it suffices to look for solutions with $ngt1$. Since $sin x$ is strictly increasing for $0le xlepi/2$, we cannot have $sin(3pi/n)=sin(5pi/n)$ if $nge10$, so it suffices to consider $2le nle9$.

From $sin x=sin(pi-x)$, we have

$$sinleft(5piover nright)=sinleft(pi-5piover nright)=sinleft((n-5)piover nright)$$

For $2le nle5$, the signs of $sin(3pi/n)$ and $sin((n-5)pi/n)$ do not agree (e.g., $sin(3pi/2)=-1$ while $sin(-3pi/2)=1$). For $6le nle9$, we have $0le3piover n,(n-5)piover nlepiover2$, in which case

$$sinleft(3piover nright)=sinleft((n-5)piover nright)iff3piover n=(n-5)piover niff3=n-5iff n=8$$

Thus we have two solutions, $n=8$ and $n=-8$.

Remark: The cases $n=3$ and $n=5$ could have been rejected outright, since $sin(3pi/n)sin(5pi/n)=0$ in those cases, guaranteeing a forbidden $0$ in one of the denominators in the original expression.