Question

Знатоки алгебры, срочно!
Найти корни уравнения
$arctg(x {}^{3} - 27x - \sqrt{3} ) = - \frac{\pi}{3}$
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Answer 1

Ответ:

x_1=0

x_2=-27^1/2

x_3=27^1/2

Объяснение:

$arctg( {x}^{3} - 27x - \sqrt{3 } ) = - \ \frac{\pi}{3}$

$- \frac{\pi}{2 } < - \frac{\pi}{3} < \frac{\pi}{2}$

tg(arctga)=a

${x}^{3} - 27x - \sqrt{3 } = tg( - \frac{\pi}{3} )$

${x}^{3} - 27x - \sqrt{3} = - \sqrt{3}$

${x}^{3} - 27x = 0$

$x( {x}^{2} - 27) = 0$

x_1=0

${x }^{2} - 27 = 0$

${x}^{2} = 27$

$x = + - \sqrt{27}$

x_2=-3×(3^1/2)

x_3=3×(3^1/2)