Решить уравнение: 1) 4cos^2 x+cos x-5=0 2) (2: 3)^8x+1=1,5^2x-3

[tex]1) 4 cos ^{2} x+cosx-5=0 \\ cosx=t \\ 4 t^{2} +t-5=0 \\ D=1+80=81 \\ \sqrt{D} =9 \\ x_{1} = \frac{-1+9}{8} =1 \\ x_{2} = \frac{-1-9}{8} =-1.25 \\ cosx=-1.25 \\ -1