Помогите решить 1) 9x^4 = (x^3-4x) ^2

2) cos^2 (2x) + cos^2 (3x) + cos^2 (10x) + cos^2 (11x) = 2

9x^4 = (x^3-4x) ^2

9x^4=x^6-8x^4+16x^2

9x^4-x^6-16x^2=0

17x^4-x^6-16x^2=0

17x^4-16x^2=x^6

x^2 (17x^2-16) = x^6

17x^2-16=x^4

x^4-17x^2+16=0 пусть x^2=t, t не равно 0

t^2-17t+16=0 t1=1 t2=16

x1=+1, x2=+-4