Question

مطلوب است محاسبه:$cos(5 \alpha - \beta )$در صورتی که: $log \frac{1}{x}-logx= \frac{4logx-1}{4logxlog \frac{1}{x} } $ ,xمثبت و $x_1= 10^{sin \alpha } $ و $x_2= 10^{sin \beta } $ $ 0 \prec \alpha \prec \beta \prec \frac{ \pi }{2} $

Answer 1

$$-2logx= \frac{4log(x)-1}{-4 (logx)^{2} } \Longrightarrow 8 (log(x))^{3}-4logx+1=0 \Longrightarrow (2log(x))^{3}-2(2logx)+1=0 \Longrightarrow 2logx=a \Longrightarrow a^{3} -2a+1=0 \Longrightarrow (a-1)( a^{2}+a-1 )=0 \Longrightarrow a=1 or a= \frac{-1+ \sqrt{5} }{2} or a= \frac{-1- \sqrt{5} }{2} \Longrightarrow 2logx=1 \Longrightarrow x= 10^{ \frac{1}{2} } ;a= \frac{-1+ \sqrt{5} }{2} \Longrightarrow 2logx= \frac{-1+ \sqrt{5} }{2} \Longrightarrow logx= \frac{-1+ \sqrt{5} }{4} \Longrightarrow x= 10^{ \frac{-1+ \sqrt{5} }{4} } \Longrightarrow \frac{-1+ \sqrt{5} }{4} \prec \frac{1}{2} \Longrightarrow sin \beta = \frac{1}{2}; \beta = \frac{ \pi }{6} ;;sin \alpha = \frac{-1+ \sqrt{5} }{4} ; \alpha = \frac{ \pi }{10} \Longrightarrow cos(5 \alpha - \beta ) =cos \frac{ \pi }{3} = \frac{1}{2} $$