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Trigonometrical Equations
hard
સમીકરણ $\sin (9 x)+\sin (3 x)=0$ ના અંતરાલ $[0,2 \pi]$ માં ઉકેલની સંખ્યા મેળવો.
A
$7$
B
$13$
C
$19$
D
$25$
(KVPY-2019)
Solution
(b)
Given trigonometric equation is
$\sin (9 x)+\sin (3 x)=0$
$\Rightarrow \quad 2 \sin 6 x \cos 3 x=0$
$\therefore \quad$ either $\sin 6 x=0$
or $\quad \cos 3 x=0$
for $x \in[0,2 \pi]$
$\sin 6 x=0$
$\Rightarrow \quad x=0, \frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{2}, \frac{2 \pi}{3}, \frac{5 \pi}{6}$,
and $3 x=0$
$\Rightarrow \quad x=\frac{\pi}{6}, \frac{\pi}{2}, \frac{5 \pi}{6}, \frac{7 \pi}{6}, \frac{3 \pi}{2}, \frac{11 \pi}{6}$
So, number of solution of given equation is $13$
Standard 11
Mathematics
