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How should I state the general solution for the equation $\sin(4\phi)=\cos(2\phi)$. The angles are $15$, $45$, $75$ and $135$ if I restrict myself within the range $[0,360]$

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As $\cos2\phi=\sin4\phi=\cos(90^\circ-4\phi)$

$\implies 2\phi=n360^\circ\pm(90^\circ-4\phi)$ where $n$ is any integer

Taking '+' sign, $2\phi=n360^\circ+90^\circ-4\phi$

$\implies 6\phi=n360^\circ+90^\circ \implies \phi=n60^\circ+15^\circ$

As $0\le \phi<360^\circ, 0\le n60^\circ+15^\circ<360^\circ\implies 0\le n\le 5$

Taking '-' sign, $2\phi=n360^\circ-90^\circ+4\phi$

$\implies 2\phi=90^\circ-n360^\circ\implies \phi=45^\circ-n180^\circ$

As $0\le \phi<360^\circ, 0\le 45^\circ-n180^\circ<360^\circ \implies -1\le n\le0$

So, there are $6+2=8$ solutions in $\in[0, 360^\circ)$

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Thanks for your effort lab bhattacharjee – DOCTOR NGILAZI BANDA JOSHUA Jul 7 '13 at 16:39
@user83845, hope I could make the idea clear – lab bhattacharjee Jul 7 '13 at 16:40

$\sin 4\phi=2\sin 2\phi\cos 2\phi=\cos 2\phi$ so either $\cos 2\phi=0$ or $\sin 2\phi=1/2$. So $2\phi = n180^\circ+90^\circ$ or $2\phi=n360^\circ+30^\circ$ or $2\phi=n360^\circ+150^\circ$

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