Toán Lớp 11: tan²x/2 – 2tanx/2 + 1 = 0 giải pt

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1. $$\begin{gathered} {\tan }^2\frac{x}{2}–2\tan \frac{x}{2}+1=0, đk: x\ne \mathrm{\pi }+\mathrm{k}2\mathrm{\pi }, \mathrm{k}\in \mathrm{Z} \\ \mathrm{Đặt} \mathrm{t}=\tan \frac{\mathrm{x}}{2} \\ =>\mathrm{Ta} \mathrm{có}: {\mathrm{t}}^2–2\mathrm{t}+1=0<=>{\left(\mathrm{t}–1\right)}^2=0<=>\mathrm{t}–1=0<=>\mathrm{t}=1 \\ \mathrm{t}=1=>\tan \frac{\mathrm{x}}{2}=1<=>\frac{\mathrm{x}}{2}=\frac{\mathrm{\pi }}{4}+\mathrm{k\pi }, \mathrm{k}\in \mathrm{Z} \\ <=>\mathrm{x}=\frac{\mathrm{\pi }}{2}+\mathrm{k}2\mathrm{\pi }, \mathrm{k}\in \mathrm{Z} \\ \mathrm{Vậy} \mathrm{S}=\left\{\frac{\mathrm{\pi }}{2}+\mathrm{k}2\mathrm{\pi }\right\}, \mathrm{k}\in \mathrm{Z} \end{gathered}$$.

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