The number of solutions of the equation sin, | vedclass

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Question

$\sin \,2x\, - 2\,\cos \,x\, + \,4\,\sin \,x\, = 4$

$ \Rightarrow \,2\,\sin \,x.\,\cos \,x - 2\,\cos \,x\, + \,4\,\sin \,x - 4 = 0$

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Vedclass · Accepted Answer

$3$

Vedclass · Answer

$\sin \,2x\, - 2\,\cos \,x\, + \,4\,\sin \,x\, = 4$

$ \Rightarrow \,2\,\sin \,x.\,\cos \,x - 2\,\cos \,x\, + \,4\,\sin \,x - 4 = 0$

$ \Rightarrow \,(\sin \,x - 1)(\cos \,x - 2)\, = \,0$

$\because \,\cos \,x\, - \,2\, 
e \,0\,,$     $	herefore \,\sin \,x\, = 1$

$	herefore \,\,x\, = \frac{\pi }{2},\frac{{5\pi }}{2},\frac{{9\pi }}{2}$

The number of solutions of the equation $sin\, 2x - 2\,cos\,x+ 4\,sin\, x\, = 4$ in the interval $[0, 5\pi ]$ is

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