![\bf 5sin(2x)=3cos(x)\qquad \qquad sin(2\theta)=2sin(\theta)cos(\theta)\qquad thus \\\\ 5[2sin(x)cos(x)]=3cos(x)\implies 10sin(x)cos(x)=3cos(x) \\\\ 10sin(x)cos(x)-3cos(x)=0\impliedby \textit{taking common factor} \\\\\\ cos(x)[10sin(x)-3]=0\implies \begin{cases} cos(x)=0\to &\measuredangle x=cos^{-1}(0) \\\\ sin(x)=\frac{3}{10}\to &\measuredangle x = sin^{-1}\left( \frac{3}{10} \right) \end{cases}](/tpl/images/0400/8563/0956a.png)
Solve using algebraic equation:
5sin2x=3cosx...
Answers: 3
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