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.