Question

4sin^2x+cos4x=1 решите плз задачу

Answer 1

4(1 - cos^2x) + 8cos^4x - 8cos^2x + 1 = 1 
4 - 4cos^2x + 8cos^4x - 8cos^2x = 0 
8cos^4x - 12cos^2x + 4 = 0   /:4
2cos^4x - 3cos^2x + 1 = 0 

Пусть cos^2x = t, тогда
2t^2  - 3t + 1 = 0 
D = 9 - 8 =  1
t1 = ( 3 + 1)/4 =  1
t2 = ( 3 -  1)/4 = 1/2

cos^2x = 1
cosx = 1
x = 2pik, k ∈Z
cosx = - 1
x = pi + 2pik, k∈Z

cos^2x = 1/2
cosx = √2/2
x = ± arccos (√2/2) + 2pik, k ∈ Z
cosx =  - √2/2
x = ± (pi - arccos √2/2) + 2pik, k ∈ Z