Solving the trigonometric equation $\sin3x=\cos2x$

Question

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.)

Answer 1

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 $x\in\left\{\frac\pi{10},\frac\pi2,\frac9{10}\pi\right\}$

Answer 2

From the triple angle identity for sine and double angle identity for cosine, the equation reduces to $$3\sin x-4\sin^3x=1-2\sin^2 x.$$ Naturally we substitute $s=\sin x$, then this becomes the cubic equation $$4s^3-2s^2-3s+1=0.$$ Now, observe that $s=1$ is a root, so factor that out from the LHS to get $$(s-1)(4s^2+2s-1)=0.$$ Solving the quadratic, we have $s=\frac14(-1\pm\sqrt5)$. Correspondingly, we obtain the $x$ values: $$x=\frac\pi2, \arcsin\left(\frac14(-1+\sqrt5)\right), \pi-\arcsin\left(\frac14(-1+\sqrt5)\right)=\frac\pi2,\frac\pi{10},\frac{9\pi}{10}$$

Answer 3

Yes, there is, using the a basic identity; rewrite the equation as $$\cos\Bigl(\frac\pi 2- 3x\Bigr)=\cos 2x\iff \frac\pi 2- 3x\equiv\pm 2x\pmod{2\pi}\iff\begin{cases} 5x\equiv \frac\pi2\\\phantom{5}x\equiv\frac\pi 2 \end{cases}\pmod{2\pi} $$ The first equation is equivalent to $$x\equiv\frac\pi{10}\pmod{\tfrac{2\pi}5}\quad\text{i.e.}\quad x=\frac{(4k+1)\pi}{10},\; k\in\mathbf Z.$$ There remains to sort the solutions which live in $[0,\pi]$.