Question

Solve using algebraic equation: 5sin2x=3cosx (No exponents)

Answer 1

$5\sin2x=3\cos x\iff10\sin x\cos x=3\cos x$

by the double angle identity for sine. Move everything to one side and factor out the cosine term.

$10\sin x\cos x=3\cos x\iff10\sin x\cos x-3\cos x=\cos x(10\sin x-3)=0$

Now the zero product property tells us that there are two cases where this is true,

$\begin{cases}\cos x=0\\10\sin x-3=0\end{cases}$

In the first equation, cosine becomes zero whenever its argument is an odd integer multiple of $\dfrac\pi2$, so $x=\dfrac{(2n+1)\pi}2$ where $n[/tex ]is any integer.\\Meanwhile,\\[tex]10\sin x-3=0\implies\sin x=\dfrac3{10}$

which occurs twice in the interval $[0,2\pi)$ for $x=\arcsin\dfrac3{10}$ and $x=\pi-\arcsin\dfrac3{10}$. More generally, if you think of $x$ as a point on the unit circle, this occurs whenever $x$ also completes a full revolution about the origin. This means for any integer $n$, the general solution in this case would be $x=\arcsin\dfrac3{10}+2n\pi$ and $x=\pi-\arcsin\dfrac3{10}+2n\pi$.