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Question Number 191512 by mnjuly1970 last updated on 25/Apr/23

$$$$$$Q:theequation$$$$$$$$\lfloor \cos \left(4x\right)\rfloor =m.\cos \left(2x\right)$$$$hasnosolution.x\in (0,\frac{\pi }{2})$$$$findtheacceptable$$$$realvaluesfor{}^{″}{m}^{″}.$$

Answered by mahdipoor last updated on 26/Apr/23

$$\left[cos\left(4x\right)\right]=0,1,-1\stackrel{noanswer}{\Rightarrow }mcos\left(2x\right)\ne 0,1,-1$$$$m\ne \frac{0,1,-1}{cos\left(2x\right)}\stackrel{1<cosx<-1}{\Rightarrow }m\ne 0,R-(-1,-1)$$$$\Rightarrow \Rightarrow m=[-1,1]$$

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