How can I solve the equation $$\sin 3x=\cos 2x$$ for $x$ in the range $[0,\pi]$?
I already have a solution elaborated in the answer below.But is there any way of solving this without using the double and triple angle identities? (The problem arises as homework in a high school mathematics class which hasn't leant the identities yet.)
Best Answer
This is $\cos(\frac\pi2-3x)=\cos(2x)$. Now, since $\cos A=\cos B\iff \exists k\in\Bbb Z, A-B=2k\pi\lor\exists h\in\Bbb Z, A+B=2h\pi$, this becomes $$\frac\pi2-3x=2k\pi+2x\lor \frac\pi2-3x=2h\pi-2x\\ x=-\frac25\pi k+\frac\pi{10}\lor x=-2h\pi+\frac\pi2$$
Now, the only such angles in $[0,\pi]$ are obtained for $k=0,-1,-2$ or for $h=0$. Therefore the solution is $x\in\left\{\frac\pi{10},\frac\pi2,\frac9{10}\pi\right\}$