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 помощью двойного неравенства:
1) -5Π/2<=2Πn<=-Π<br>-5Π/4<=Πn<=-Π/2<br>-5/4<=n<=-1/2<br>n=-1
x=2Π*(-1)=-2Π
2) -5Π/2<=-2Π/3+2Πk<=-Π<br>-5Π/2+2Π/3<=2Πk<=-Π+2Π/3<br>-11Π/6<=2Πk<=-Π/3<br>-11Π/12<=Πk<=-Π/6<br>-11/12<=k<=-1/6<br>3) -5Π/2<=2Π/3+2Πk<=-Π<br>-5Π/2-2Π/3<=2Πk<=-Π-2Π/3<br>-19Π/6<=2Πk<=-5Π/3<br>-19Π/12<=Πk<=-5Π/6<br>-19/12<=k<=-5/6<br>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.