Question

Расписать решение
1.Решить уравнение(выбрать корни)
2.Решить неравенство

Answer 1

$1) a) 3mathrm{tg}x - 2mathrm{ctg}x - 1= 0 \ \ 3dfrac{sin x}{cos x} - 2dfrac{cos x}{sin x} - 1= 0 / * mathrm{tg}x ne 0 \ \ 3mathrm{tg^{2}}x - 2 - mathrm{tg}x = 0 \ \ mathrm{tg}x = t \ \ 3t^{2} - t - 2 = 0 \ D = 1 + 24 = 25 \ \ t_{1} = dfrac{1+5}{6} = 1 ; t_{2} = -dfrac{2}{3} \ \ left[ begin{gathered} mathrm{tg}x = 1 \ mathrm{tg}x = -dfrac{2}{3} \ end{gathered} right.$

$left[ begin{gathered} x = dfrac{pi}{4} + pi n, n in Z \ x = -mathrm{arctg}dfrac{2}{3} + pi k, k in Z end{gathered} right.$

$b) pi + dfrac{pi}{4} = dfrac{5pi}{4}$

Ответ: a) π/4 + πn, n ∈ Z, arctg(-2/3) + pi k, k in Z

b) 5π/4, arctg(-2/3) + π/2, arctg(-2/3) + 3π/2

$2) sqrt{1-9log ^{2}_{8}x} - 4log_{8}x > 1 \ \ ODZ: left{ begin{gathered} x > 0 \ 1-9log_{8}^{2}x ge 0 (1) \ end{gathered} right.$

$(1): 1 - 9log_{8}^{2}x ge 0 \ \ 9log_{8}^{2}x le 1 \ \ log_{8}^{2}x le dfrac{1}{9} \ \ |log_{8}x| le dfrac{1}{3} \ \ -dfrac{1}{3} le log_{8}x le dfrac{1}{3} \ \ log_{8}8^{-frac{1}{3}} le log_{8}x le log_{8}8^{frac{1}{3}} \ \ 8^{-frac{1}{3}} le x le 8^{frac{1}{3}} \ \ dfrac{1}{2} le x le 2 ; x in [dfrac{1}{2};2]$

$x in [dfrac{1}{2} ;2]$

Решим уравнение:

$sqrt{1-9log ^{2}_{8}x} > 1 + 4log_{8}x \ \ sqrt{1-9log_{2^{3}}^{2}x} > 1 + 4log_{2^{3}}x \ \ sqrt{1 -log_{2}^{2}x} > 1 + dfrac{4}{3}log_{2}x \ \ (sqrt{1 -log_{2}^{2}x})^{2} > (1 + dfrac{4}{3}log_{2}x)^{2}$

$left{ begin{gathered} 1 - log_{2}^{2}x textgreater 1 + dfrac{8}{3}log_{2}x + dfrac{16}{9}log_{2}^{2}x (2) \ 1 + dfrac{4}{3}log_{2}x ge 0 (3)\ end{gathered}$
ИЛИ
$left{ begin{gathered} 1 - log_{2}^{2}x textgreater -(1 + dfrac{8}{3}log_{2}x + dfrac{16}{9}log_{2}^{2}x) (3) \ 1 + dfrac{4}{3}log_{2}x textless 0 (4) \ end{gathered} right.$ 

$(2): 1 - log_{2}^{2}x > 1 + dfrac{8}{3}log_{2}x + dfrac{16}{9}log_{2}^{2}x \ \ log_{2}x = t \ \ 1 - t^{2} > 1 + dfrac{8t}{3} + dfrac{16t^{2}}{9} \ \ 9 - 9t^{2} > 9 + 24t + 16t^{2} \ \ -25t^{2} - 24t > 0 \ \ -t(25t + 24) > 0 left{ begin{gathered} t textless 0 \ t textgreater -dfrac{24}{25} \ end{gathered} right.$

$t in (-dfrac{24}{25}; 0)$

$-dfrac{24}{25} < log_{2}x < 0 \ \ log_{2}2^{-frac{24}{25}} < log_{2}x < log_{2}2^{0} \ \ 2^{-frac{24}{25}}< x < 1 ; x in (dfrac{1}{sqrt[25]{2^{24}}}; 1)$

$(3): 1 + dfrac{4}{3}log_{2}x ge 0 \ \ log_{2}x ge log_{2}2^{-frac{3}{4}} \ \ x ge 2^{-frac{3}{4}} ; x in [dfrac{1}{sqrt[4]{2^{3}}}; +infty)$

Пересечём (2) и (3):

$x in [dfrac{1}{sqrt[4]{2^{3}}};1)$

$(3): x in R \ \ (4): 1 + dfrac{4}{3}log_{2}x textless 0 \ \ log_{2}x textless - dfrac{3}{4} \ \ x textless dfrac{1}{ sqrt[4]{2^{3}}}$


Пересечём (2) и (3) с (3) и (4):

$x in (-infty; 1)$

Учтём ОДЗ (рисунок 3):

$x in [dfrac{1}{2}; 1)$

Ответ: x ∈ [1/2; 1)