2*16^(cosx) -9*4^(cosx) +4=0 Найти корни в промежутке [-3П;-3П/2]. Created by MCscope. algebra-ru

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

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paradiseva

Verified answer

$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:
$x_1= - \frac{8 \pi }{3}\\x_2=-2 \pi$

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