Question

Решить уравнение:
2√2 cosx+2-cos2x=0

Answer 1

 2√2cosx + 2 - (2cos²x - 1) = 0
2cos²x - 2√2cosx - 3 = 0
cosx = t
2t² - 2√2t - 3 = 0
D/4 = (b/2)² - ac = 2 + 6 = 8
t = (-b/2 +- √(D/4)) / a
t = (√2 - 2√2)/2 = - √2/2                t = (√2 + 2√2)/2 = 3√2/2

cosx = - √2/2                                  cosx = 3√2 / 2 > 1 нет корней
x = + - arccos (- √2/2) + 2πn
x = + - (π - arccos(√2/2)) + 2πn
x = + - 3π/4 + 2πn,      n∈Z