Помогите 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=0Пусть t=cosx, где t€[-1;1], тогда-4t^2+2t+2=0-2t^2+t+1=0D=1+8=9t1=-1-3/-4=1t2=-1+3/-4=-1/2Вернёмся к замене:cosx=1x=2Πn, n€Zcosx=-1/2x=-2Π/3+2Πk, k€Zx=2Π/3+2Πk, k€Zб) Решим с помощью двойного неравенства:1) -5Π/2<=2Πn<=-Π-5Π/4<=Πn<=-Π/2-5/4<=n<=-1/2n=-1x=2Π*(-1)=-2Π2) -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<=-1/63) -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<=-5/6k=-1x=2Π/3+2Π(-1)=2Π/3-2Π=-4Π/3Ответ: а) 2Πn, n€Z; +-2Π/3+2Πk, k€Z, б) -2Π; -4Π/3.