Решите уравнение при [tex]1 \leq x \leq 6[/tex]
[tex]3log^2_{2}sin{\frac{ \pi x}{3}}+log_{2}\big(1-cos{\frac{2 \pi x}{3}}\big)=2[/tex]
[tex]3log^2_{2}sin{\frac{ \pi x}{3}}+log_{2}\big(1-cos{\frac{2 \pi x}{3}}\big)=2[/tex]
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1-cos2πx/3=sin²πx/3+cos²πx/3-cos²πx/3+sin²πx/3=2sin²πx/3log(2)(2sin²πx/3)=1+2log(2)sinπx/3
3log²(2)sinπx/3+1+2log(2)sinπx/3-2=0
log(2)sinπx/3=a
3a²+2a-1=0
D=4+12=16
a1=(-2-4)/6=-1⇒log(2)sinπx/3=-1
sinπx/3=1/2
πx/3=π/6+2πk U πx/3=5π/6+2πk
x=1/2++6k,k∈z U x=5/2+6k
1≤1/2+6k≤6 U 1≤5/2+6k≤6
2≤1+12k≤12 U 2≤5+12k≤12
1≤12k≤11 U -3≤12k≤7
1/12≤k≤11/12 нет решения U -1/4≤k≤7/12⇒k=0⇒x=5/2
a2=(-2+4)/6=1/3⇒log(2)sinπx/3=1/3
sinπx/3=∛2>1 нет решения
Ответ x=2,5