Question

2*16^(cosx) -9*4^(cosx) +4=0 найти корни в промежутке [-3п; -3п/2]

Answer 1

$2*16^{cosx}-9*4^{cosx}+4=0\\ 4^{cosx}=t, t\ \textgreater \ 0\\ 2t^2-9t+4=0\\ D=81-4*2*4=49\\ t_1= \frac{9+7}{4}=4\\ t_2= \frac{9-7}{4}= \frac{1}{2} \\ 4^{cosx}=4\\ cosx=1\\ x=2 \pi k, k \in Z\\ k=-1 \Rightarrow x=-2 \pi \in [-3 \pi ;- \frac{3 \pi }{2} ]\\ k=-2 \Rightarrow x=-4 \pi \notin [-3 \pi ;- \frac{3 \pi }{2} ]\\$
$4^{cosx}= \frac{1}{2} \\ 2^{2cosx}= 2^{-1}\\ 2cosx=-1\\ cosx= -\frac{1}{2} \\ x = \pm arccos (-\frac{1}{2} )+2 \pi n, n \in Z\\ x = \pm ( \pi - arccos \frac{1}{2})+2 \pi n, n \in Z\\ x = \pm ( \pi - \frac{ \pi }{3})+2 \pi n, n \in Z\\$
$x = \pm \frac{ 2\pi }{3}+2 \pi n, n \in Z\\ n=0 \Rightarrow \\ x_1 = \frac{ 2\pi }{3} \notin [-3 \pi ;- \frac{3 \pi }{2} ], x_2= -\frac{ 2\pi }{3} \notin [-3 \pi ;- \frac{3 \pi }{2} ]\\ n=-1 \Rightarrow \\ x_1 = \frac{ 2\pi }{3}-2 \pi =- \frac{4 \pi }{3} \notin [-3 \pi ;- \frac{3 \pi }{2} ], \\x_2= -\frac{ 2\pi }{3}-2 \pi =- \frac{8 \pi }{3} \in [-3 \pi ;- \frac{3 \pi }{2} ]\\$
$n=-2 \Rightarrow \\ x_1 = \frac{ 2\pi }{3}-4 \pi =- \frac{10 \pi }{3} \notin [-3 \pi ;- \frac{3 \pi }{2} ], \\ x_12 = -\frac{ 2\pi }{3}-4 \pi =- \frac{14 \pi }{3} \notin [-3 \pi ;- \frac{3 \pi }{2} ],$
ответ:
$x_1= - \frac{8 \pi }{3}\\x_2=-2 \pi$