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Values of x for 2 \sin x \leq sin\ 2x in x\epsilon [0,\frac{\pi}{2}]

Option: 1

x\epsilon \phi


Option: 2

x\epsilon [0,\frac{\pi}{2}]


Option: 3

x\epsilon [0,\frac{\pi}{4}]


Option: 4

x\epsilon [\frac{\pi}{4},\frac{\pi}{2}]


Answers (1)

best_answer

\\2 \sin x \leq \sin\ 2x\\ \text{For intersection point } \\2\sin x= \sin 2x\\ \sin 2x -2\sin x=0\\ 2\sin x \cos x-2\sin x=0\\ 2 \sin x (\cos x-1)=0\\ \sin x= 0\ or\ \cos x=1 \\ x=0

(As we have to look for x lying in [0,\pi/2] only)

Graph 

By graph, we can say that no solution for this inequalities 

Posted by

Nehul

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