Question

Arcosx=arctgx найти корень уравнения

Answer 1

$\sqrt{\frac{\sqrt{5}-1}{2}}$

Пошаговое объяснение:

$acrrosx=arctgx\\\\tg(arccosx)=tg(arctgx)\\\\\frac{\sqrt{1-x^2}}{x}=x,\; \; \; x\neq 0\\\\\sqrt{1-x^2}=x^2\\\\(\sqrt{1-x^2})^2=(x^2)^2\\\\1-x^2=x^4\\\\x^4+x^2-1=0\\\\y=x^2,\; \; \; x\neq 0=y0\\\\y^2+y-1=0\\\\D=1^2-4*1*(-1)=1+4=5\\\\y_1=\frac{-1+\sqrt{5}}{2}=\frac{\sqrt{5}-1}{2} \; (0)\; \; \; \; y_2=\frac{-1-\sqrt{5}}{2}\; (x=\sqrt{y}\\\\x=\sqrt{\frac{\sqrt{5}-1}{2}}$

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$tg(arctgx)=x,\; x\in R\\\\tgx=\frac{\sqrt{1-cos^2x}}{cosx},\; x\in[0;\pi/2)\cup (\pi/2;\pi]\; = \\\\=tg(arccosx)=\frac{\sqrt{1-cos^2(arccosx)}}{cos(arccosx)}= \frac{\sqrt{1-x^2}}{x},\; x\in(-1;0)\cup (0;1)$