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Trigonometrical Equations
medium
The number of real solutions of the equation $2 \sin 3 x+\sin 7 x-3=0$, which lie in the interval $[-2 \pi, 2 \pi]$ is
A
$1$
B
$2$
C
$3$
D
$4$
(KVPY-2017)
Solution
(b)
Given, $2 \sin 3 x+\sin 7 x-3=0$
$\Rightarrow \quad 2 \sin 3 x+\sin 7 x=3$
lt is possible only $\sin 3 x=1$ and $\sin 7 x=1$
$\therefore \quad \sin 3 x =1$
$\therefore \quad x =n \pi+(-1)^n \frac{\pi}{2}$
$x =\frac{n \pi}{3}+(-1)^n \frac{\pi}{6}$
$\because \quad \sin 7 x =1$
$\therefore \quad 7 x =m \pi+(-1)^m \frac{\pi}{2}$
$x =\frac{m \pi}{7}+(-1)^m \frac{\pi}{14}$
Hence, only two solution exists.
Standard 11
Mathematics
