Question

Помогите, пожалуйста, решить уравнение:

cos²2x-sin²x =0

Answer 1

cos²2x-sin²x =0
(1+сos4x)/2-(1-cos2x)/2=0
1+cos4x-1+cos2x=0
cos4x+cos2x=0
2cos²2x-1+cos2x=0
cosx=t
2t²+t-1=0
D=1+8=9
t1=(-1-3)/4=-1⇒cos2x=-1⇒2x=π+2πn⇒x=π/2+πn,n∈z
t2=(-1+3)/4=1/2⇒cos2x=1/2⇒2x=+-π/3+2πn⇒x=+-π/6+πn,n∈z

Answer 2

(1+cos4x)/2 - (1-cos2x)/2=0

(1+cos4x-1+cos2x)/2=0

co4x+cos2x=0

2cos²2x -1+cos2x=0

cos2x=a
2a²+a-1=0
D=1+8=9
a1=-1+3/4=2/4=1/2
a2=-1-3/4=-4/4=-1

cos2x=1/2
2x=+- π/3+2πk
x=+-π/6+πk

cos2x=-1
2x=π+2πk
x=π/2+πk