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
`\frac\pi2`, so
`x=\frac{(2n+1)\pi}2` where
`image`

which occurs twice in the interval
`[0,2\pi)` for
`x=\arcsin\frac3{10}` and
`x=\pi-\arcsin\frac3{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\frac3{10}+2n\pi` and
`x=\pi-\arcsin\frac3{10}+2n\pi`.