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atroni [7]
3 years ago
10

Solve using algebraic equation: 5sin2x=3cosx

Mathematics
2 answers:
disa [49]3 years ago
5 0
\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}
Mila [183]3 years ago
4 0

Answer:

x = 17.46° or x = 90°

Step-by-step explanation:

We need to solve 5sin2x=3cosx

sin2x =

Substituting

5\times 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°

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