Question

Solve using algebraic equation:
5sin2x=3cosx

Answer 1

`\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)=(3)/(10)\to &\measuredangle x = sin^(-1)\left( (3)/(10) \right) \end{cases}`

Answer 2

x = 17.46° or x = 90°

Explanation:

We need to solve 5sin2x=3cosx

sin2x =

Substituting

`5* 2sinx cosx = 3 cosx\\\\cosx(10sinx-3)=0\\\\sinx=0.3\texttt{ or }cosx=0\\\\x=17.46^0\texttt{ or }x=90^0`

x = 17.46° or x = 90°