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Question Number 188170 by mnjuly1970 last updated on 26/Feb/23

$$$$$$solve$$$$$$$$\lfloor cos\left(x\right)\rfloor +\lfloor cos\left(x\right)+\frac{1}{2}\rfloor +\lfloor -2cosx\rfloor =0$$$$$$

Answered by Frix last updated on 26/Feb/23

$$\mathrm{Let}f\left(t\right)=\lfloor t\rfloor +\lfloor t+\frac{1}{2}\rfloor +\lfloor -2t\rfloor$$$$\mathrm{Let}\mathbb{T}=\{\frac{n}{2}\mid n\in \mathbb{Z}\}$$$${\mathrm{It}}^{\prime }\mathrm{s}\mathrm{easy}\mathrm{to}\mathrm{see}$$$$f\left(t\right)=\{\begin{array}{l}0;t\in \mathbb{T}\\ -1;t\in \mathbb{R}\mathrm{\setminus }\mathbb{T}\end{array}$$$$\mathrm{Let}t=\cos x\Rightarrow -1⩽t<1\Rightarrow$$$$f\left(t\right)=0\iff t=\cos x\in \{-1,-\frac{1}{2},0,\frac{1}{2},1\}$$$$\mathrm{The}\mathrm{rest}\mathrm{is}\mathrm{easy}$$

Commented by mnjuly1970 last updated on 27/Feb/23

$$thxalot$$$$tidomebonsir...$$

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