Menu Close

solve-cos-x-cos-x-1-2-2cosx-0-




Question Number 188170 by mnjuly1970 last updated on 26/Feb/23
        solve     ⌊ cos (x )⌋ + ⌊ cos(x) +(1/2) ⌋+ ⌊ −2cosx ⌋ =0
$$ \\ $$$$\:\:\:\:\:\:{solve} \\ $$$$ \\ $$$$\:\lfloor\:{cos}\:\left({x}\:\right)\rfloor\:+\:\lfloor\:{cos}\left({x}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}\:\rfloor+\:\lfloor\:−\mathrm{2}{cosx}\:\rfloor\:=\mathrm{0} \\ $$$$ \\ $$
Answered by Frix last updated on 26/Feb/23
Let f(t)=⌊t⌋+⌊t+(1/2)⌋+⌊−2t⌋  Let T={(n/2)∣n∈Z}  It′s easy to see  f(t)= { ((0; t∈T)),((−1; t∈R\T)) :}  Let t=cos x ⇒ −1≤t<1 ⇒  f(t)=0 ⇔ t=cos x ∈{−1, −(1/2), 0, (1/2), 1}  The rest is easy
$$\mathrm{Let}\:{f}\left({t}\right)=\lfloor{t}\rfloor+\lfloor{t}+\frac{\mathrm{1}}{\mathrm{2}}\rfloor+\lfloor−\mathrm{2}{t}\rfloor \\ $$$$\mathrm{Let}\:\mathbb{T}=\left\{\frac{{n}}{\mathrm{2}}\mid{n}\in\mathbb{Z}\right\} \\ $$$$\mathrm{It}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{see} \\ $$$${f}\left({t}\right)=\begin{cases}{\mathrm{0};\:{t}\in\mathbb{T}}\\{−\mathrm{1};\:{t}\in\mathbb{R}\backslash\mathbb{T}}\end{cases} \\ $$$$\mathrm{Let}\:{t}=\mathrm{cos}\:{x}\:\Rightarrow\:−\mathrm{1}\leqslant{t}<\mathrm{1}\:\Rightarrow \\ $$$${f}\left({t}\right)=\mathrm{0}\:\Leftrightarrow\:{t}=\mathrm{cos}\:{x}\:\in\left\{−\mathrm{1},\:−\frac{\mathrm{1}}{\mathrm{2}},\:\mathrm{0},\:\frac{\mathrm{1}}{\mathrm{2}},\:\mathrm{1}\right\} \\ $$$$\mathrm{The}\:\mathrm{rest}\:\mathrm{is}\:\mathrm{easy} \\ $$
Commented by mnjuly1970 last updated on 27/Feb/23
thx alot   tidomebon sir...
$${thx}\:{alot}\: \\ $$$${tidomebon}\:{sir}… \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *