Solve $\sin 2 x-\sin 4 x+\sin 6 x=0$
Solve $\sin 2 x-\sin 4 x+\sin 6 x=0$
The equation can be written as
$\sin 6 x+\sin 2 x-\sin 4 x=0$
or $2 \sin 4 x \cos 2 x-\sin 4 x=0$
i.e. $\quad \sin 4 x(2 \cos 2 x-1)=0$
Therefore $\sin 4 x=0 \quad$ or $\cos 2 x=\frac{1}{2}$
i.e. $\sin 4 x=0$ or $\cos 2 x=\cos \frac{\pi}{3}$
Hence $\quad 4 x=n \pi$ or $2 x=2 n \pi \pm \frac{\pi}{3},$ where $n \in Z$
i.e. $x=\frac{n \pi}{4}$ or $x=n \pi \pm \frac{\pi}{6},$ where $n \in Z$
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If $a = \sin \frac{\pi }{{18}}\sin \frac{{5\pi }}{{18}}\sin \frac{{7\pi }}{{18}}$ and $x$ is the solution of the equatioin $y = 2\left[ x \right] + 2$ and $y = 3\left[ {x - 2} \right] ,$ where $\left[ x \right]$ denotes the integral part of $x,$ then $a$ is equal to :-
If $a = \sin \frac{\pi }{{18}}\sin \frac{{5\pi }}{{18}}\sin \frac{{7\pi }}{{18}}$ and $x$ is the solution of the equatioin $y = 2\left[ x \right] + 2$ and $y = 3\left[ {x - 2} \right] ,$ where $\left[ x \right]$ denotes the integral part of $x,$ then $a$ is equal to :-
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Statement $-2:$ The number of solutions of the equation, $2\,cos^2\,\theta - 3\,sin\,\theta = 0$ in the interval $[0, \pi ]$ is two.
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Statement $-2:$ The number of solutions of the equation, $2\,cos^2\,\theta - 3\,sin\,\theta = 0$ in the interval $[0, \pi ]$ is two.
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