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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?

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  • $\begingroup$ Hint: $\sin(x) = \cos(\pi /2 - x) $ and Prosthaphaeresis Formulas: $ \sin \alpha - \sin \beta = \ldots $ $\endgroup$ – GohP.iHan Jan 19 '16 at 7:32
  • $\begingroup$ @GohP.iHan , I'm afraid I cannot use those.. (school does not allow it) $\endgroup$ – bigfocalchord Jan 19 '16 at 7:33
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    $\begingroup$ you fifth line has an error. It should have $-2\sin^3 x.$ $\endgroup$ – Anurag A Jan 19 '16 at 7:38
  • $\begingroup$ @AnuragA oh thanks! $\endgroup$ – bigfocalchord Jan 19 '16 at 7:38
  • $\begingroup$ sin x=1 is one solution. $\endgroup$ – DanielWainfleet Jan 19 '16 at 7:41
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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.

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$$\cos2x=\sin3x=\cos\left(\dfrac\pi2-3x\right)$$

$$\iff2x=2m\pi\pm\left(\dfrac\pi2-3x\right)$$ where $m$ is any integer

Alternatively, $$\sin3x=\cos2x=\sin\left(\dfrac\pi2-2x\right)$$

$$3x=n\pi+(-1)^n\left(\dfrac\pi2-2x\right)$$ where $n$ is any integer

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You have made some errors in your calculations (or some typos here).

$$\sin(3x)=\cos(2x)$$

$$ \sin(2x+x) = \cos(2x)$$

$$\sin(2x)\cos(x) + \cos(2x)sin(x) = \cos(2x) $$

$$ 2\sin(x)\cos(x)\cos(x) + (1-2\sin^2(x))\sin(x)) = \cos(2x) $$

$$ 2\sin(x)\cos^2(x) + \sin(x) - 2\sin^{\bf{3}}(x) = \cos(2x) $$

$$ 2\sin(x)(1-\sin^2(x))+\sin(x)-2\sin^{\bf{3}}(x)=\cos(2x) $$

$$ 2\sin(x) - 2\sin^3(x) + \sin(x)- 2 \sin^{\bf{3}}(x) = \cos(2x) $$

$$3\sin(x)-4\sin^3(x)=\cos(2x)$$

Then recall that $\cos(2x)=1-2\sin^2(x)$ to give:

$$3\sin(x)-4\sin^3(x)=1-2\sin^2(x)$$

This is a cubic in $\sin(x)$. For simplicity write $y=\sin(x)$ to get:

$$-4y^3+2y^2+3y-1=0$$

$$-(y-1)(4y^2+2y-1)=0$$

So $\sin(x)=1$ or $\sin(x)=\frac{-1\pm\sqrt{5}}{4}$

So $x=\frac{\pi}{2}$ or $x=\frac{\pi}{10}$, $\frac{9\pi}{10}$, $\frac{13\pi}{10}$, $\frac{17\pi}{10}$

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