Question

Решить уравнение при $2\leq x < 3.5$
$2 \log_{5}\|ctg\frac{ \pi x}{2}\|-log_{\frac15}\frac{sin(\frac{ \pi x}{2})}{5sin(\frac{ \pi x}{2})-4cos(\frac{ \pi x}{2})}=0$

Answer 1

$2\log_{5} \left|\text{ctg} \dfrac{\pi x}{2} \right| - \log_{\frac{1}{5} } \dfrac{\sin \dfrac{\pi x}{2} }{5 \sin \dfrac{\pi x}{2} - 4 \cos \dfrac{\pi x}{2}} = 0$

$\log_{5} \left|\text{ctg} \dfrac{\pi x}{2} \right|^{2} - \log_{5^{-1}} \left( \dfrac{5 \sin \dfrac{\pi x}{2} - 4 \cos \dfrac{\pi x}{2}}{\sin \dfrac{\pi x}{2}} \right)^{-1} = 0$

$\log_{5} \text{ctg}^{2} \dfrac{\pi x}{2} - \log_{5} \dfrac{5 \sin \dfrac{\pi x}{2} - 4 \cos \dfrac{\pi x}{2}}{\sin \dfrac{\pi x}{2}} = 0$

$\log_{5} \text{ctg}^{2} \dfrac{\pi x}{2} - \log_{5} \left( \dfrac{5 \sin \dfrac{\pi x}{2}}{\sin \dfrac{\pi x}{2}} - \dfrac{4\cos \dfrac{\pi x}{2}}{\sin \dfrac{\pi x}{2}} \right) = 0$

$\log_{5} \text{ctg}^{2} \dfrac{\pi x}{2} = \log_{5} \left( 5 - {4\text{ctg} \dfrac{\pi x}{2} \right)$

$\text{ctg}^{2} \dfrac{\pi x}{2} =5 - {4\text{ctg} \dfrac{\pi x}{2}$

$\text{ctg}^{2} \dfrac{\pi x}{2} + {4\text{ctg} \dfrac{\pi x}{2} - 5 = 0$

Замена: $\text{ctg} \dfrac{\pi x}{2} = t$

$t^{2} + 4t - 5 = 0$

$\displaystyle \left [ {{t_{1} = -5} \atop {t_{2} = 1 \ \ }} \right.$

Обратная замена:

$\displaystyle \left [ {{\text{ctg} \dfrac{\pi x}{2}= -5} \atop {\text{ctg} \dfrac{\pi x}{2} = 1 \ \ }} \right.$

$\displaystyle \left [ {{\dfrac{\pi x}{2}= \text{arcctg}(-5) + \pi n} \atop {\dfrac{\pi x}{2}= \text{arcctg} \, 1 + \pi n \ \ \ \ }} \right. \ \ \ n \in Z$

$\displaystyle \left [ {{\dfrac{\pi x}{2}= \pi - \text{arcctg}(5) + \pi n} \atop {\dfrac{\pi x}{2}= \dfrac{\pi}{4} + \pi n \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }} \right. \ \ \ n \in Z$

$\displaystyle \left [ {{x = 2 - \dfrac{2\,\text{arcctg}(-5)}{\pi} + 2n} \atop {x= \dfrac{1}{2} + 2n \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }} \right. \ \ \ n \in Z$

Если $n = 1$, то $\displaystyle \left [ {{x = 4 -\dfrac{2\,\text{arcctg}(-5)}{\pi} < 2,5} \atop {x= 2,5 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }} \right.$

Ответ: $x = 2,5$