Решите уравнение √2cos^2x=sinx
Решите уравнение √2cos^2x=sinx

V2*cos^2 x = sin xV2 * (1 - sin^2 x) = sin xV2*sin^2 x + sin x - V2 = 0sin x = tV2*t + t - V2 = 0 ------> t1=V2; t2=V2\2Дальше сам

√2cos2x = cosx+sinx√2(cos²x - sin²x) - (cosx + sinx) = 0√2(sinx + cosx)(cosx - sinx) - (cosx + sinx) = 0(sinx + cosx)(√2cosx - √2sinx - 1) = 01) sinx + cosx = 0sinx = -cosxtgx = -1x = -π/4 + πn, n ∈ Z2) √2cosx - √2sinx - 1 = 0√2cosx - √2sinx = 1√2/2cosx - √2/2sinx = 1/2cosx·cos(arccos(√2/2) - sinx·sin(arccos(√2/2)) = 1/2cos(x + arccos(√2/2)) = 1/2cosx(x + π/4) = 1/2x + π/4 = ±π/3 + 2πk, k ∈ Zx = ± π/3 - π/4 + 2πk, k ∈ ZОтвет: x = -π/4 + πn, n ∈ Z; ± π/3 - π/4 + 2πk, k ∈ Z.Подробнее - на Znanija.com - https://znanija.com/task/25223827#readmore