Question: Solve $\sin(3x)=\cos(2x)$ for $0≤x≤2\pi$.
My knowledge on the subject; I know the general identities, compound angle formulas and double angle formulas so I can only apply those.
With that in mind
\begin{align}
\cos(2x)=&~ \sin(3x)\\
\cos(2x)=&~ \sin(2x+x) \\
\cos(2x)=&~ \sin(2x)\cos(x) + \cos(2x)\sin(x)\\
\cos(2x)=&~ 2\sin(x)\cos(x)\cos(x) + \big(1-2\sin^2(x)\big)\sin(x)\\
\cos(2x)=&~ 2\sin(x)\cos^2(x) + \sin(x) – 2\sin^2(x)\\
\cos(2x)=&~ 2\sin(x)\big(1-\sin^2(x)\big)+\sin(x)-2\sin^2(x)\\
\cos(2x)=&~ 2\sin(x) – 2\sin^3(x) + \sin(x)- 2 \sin^2(x)\\
\end{align}
edit
\begin{gather}
2\sin(x) – 2\sin^3(x) + \sin(x)- 2 \sin^2(x) = 1-2\sin^2(x) \\
2\sin^3(x) – 3\sin(x) + 1 = 0
\end{gather}
This is a cubic right?
So $u = \sin(x)$,
\begin{gather} 2u^3 – 3u + 1 = 0 \\
(2u^2 + 2u – 1)(u-1) = 0
\end{gather}
Am I on the right track?
This is where I am stuck what should I do now?
Best Answer
Use $\sin 3x=3 \sin x - 4 \sin^3x$ and $\cos 2x=1-2\sin^2x$. To get $$3 \sin x - 4 \sin^3x=1-2\sin^2x.$$ Now call $\sin x=t$. Thus we have $$4t^3-2t^2-3t+1=0.$$ Observe that $t=1$ is definitely a solution, so we have $$(t-1)(4t^2+2t-1)=0.$$ The quadratic factor will be zero for $$t=\frac{-1\pm \sqrt{5}}{4}$$ I hope you can solve from here.