Решить уравнение √2 cos2x = cosx+sinx

√2cos2x = cosx+sinx
√2(cos²x - sin²x) - (cosx + sinx) = 0
√2(sinx + cosx)(cosx - sinx) - (cosx + sinx) = 0
(sinx + cosx)(√2cosx - √2sinx - 1) = 0
1) sinx + cosx = 0
sinx = -cosx
tgx = -1
x = -π/4 + πn, n ∈ Z
2) √2cosx - √2sinx - 1 = 0
√2cosx - √2sinx  = 1
√2/2cosx - √2/2sinx = 1/2
cosx·cos(arccos(√2/2) - sinx·sin(arccos(√2/2)) = 1/2
cos(x + arccos(√2/2)) = 1/2
cosx(x + π/4) = 1/2
x + π/4  = ±π/3 + 2πk, k ∈ Z
x =  ± π/3 - π/4 + 2πk, k ∈ Z
ответ: x = -π/4 + πn, n ∈ Z; ± π/3 - π/4 + 2πk, k ∈ Z.