Решиить и найти значение в промежутке [-5п/2; -п] 1/49^cos2x=7^2-2cosx

А) (1/49)^cos2x=7^(2-2cosx)
7^-2(cos2x)=7^(2-2cosx)
7^-2cos2x=7^(2-2cosx)
Логарифмируя показательное ур-ие, получим ур-ие:
-2cos2x=2-2cosx
-2cos2x-2+2cosx=0
-2(cos2x)+2cosx-2=0
-4(cos^2x-1)+2cosx-2=0
-4cos^2x+2cosx+2=0
Пусть t=cosx, где t€[-1;1], тогда
-4t^2+2t+2=0
-2t^2+t+1=0
D=1+8=9
t1=-1-3/-4=1
t2=-1+3/-4=-1/2
Вернёмся к замене:
cosx=1
x=2Πn, n€Z
cosx=-1/2
x=-2Π/3+2Πk, k€Z
x=2Π/3+2Πk, k€Z
б) Решим с двойного неравенства:
1) -5Π/2<=2Πn<=-Π
-5Π/4<=Πn<=-Π/2
-5/4<=n<=-1/2
n=-1
x=2Π*(-1)=-2Π
2) -5Π/2<=-2Π/3+2Πk<=-Π
-5Π/2+2Π/3<=2Πk<=-Π+2Π/3
-11Π/6<=2Πk<=-Π/3
-11Π/12<=Πk<=-Π/6
-11/12<=k<=-1/6
3) -5Π/2<=2Π/3+2Πk<=-Π
-5Π/2-2Π/3<=2Πk<=-Π-2Π/3
-19Π/6<=2Πk<=-5Π/3
-19Π/12<=Πk<=-5Π/6
-19/12<=k<=-5/6
k=-1
x=2Π/3+2Π(-1)=2Π/3-2Π=-4Π/3
ответ: а) 2Πn, n€Z; +-2Π/3+2Πk, k€Z, б) -2Π; -4Π/3.