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The number of real solutions of the equation 2sin $3 x+\sin 7 x-3=0$ which lie in the interval $[-2 \pi, 2 \pi]$ is
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Verified Answer
The correct answer is:
2
only possible when $\sin 3 x=1 \& \sin 7 x=1$ $\sin 3 x=1$
$\begin{array}{l}
\sin 3 x=\sin (4 n+1) \frac{\pi}{2}, n \in I \\
3 x=(4 n+1) \frac{\pi}{2} \Rightarrow x=(4 n+1) \frac{\pi}{6} \\
\sin 7 x=\sin (4 m+1) \frac{\pi}{2}, m \in I \\
x=(4 m+1) \frac{\pi}{14}
\end{array}$
for common solution
$(4 n+1) \frac{\pi}{6}=(4 m+1) \frac{\pi}{14}$
Solving these $1=3 \mathrm{~m}-7 \mathrm{n}$
First solution is $m=5, n=2$
Second solution is $\mathrm{m}=12, \mathrm{n}=5$
So two solutions are possible
$\begin{array}{l}
\sin 3 x=\sin (4 n+1) \frac{\pi}{2}, n \in I \\
3 x=(4 n+1) \frac{\pi}{2} \Rightarrow x=(4 n+1) \frac{\pi}{6} \\
\sin 7 x=\sin (4 m+1) \frac{\pi}{2}, m \in I \\
x=(4 m+1) \frac{\pi}{14}
\end{array}$
for common solution
$(4 n+1) \frac{\pi}{6}=(4 m+1) \frac{\pi}{14}$
Solving these $1=3 \mathrm{~m}-7 \mathrm{n}$
First solution is $m=5, n=2$
Second solution is $\mathrm{m}=12, \mathrm{n}=5$
So two solutions are possible
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