Log_x+19 (2x^2+36x+1) = log_4 (8) + cos^2 (5pi/4) ;

Log_x+19 (2x^2+36x+1) = log_4 (8) + cos^2 (5pi/4)

Log_x+19 (2x^2+36x+1) = log_4 (8) + 1/2

Log_x+19 (2x^2+36x+1) = 3/2+1/2

Log_x+19 (2x^2+36x+1) = l2

ОДЗ

{x+19>0 ⇒x>-19

{x+19≠1⇒x≠-18

{2x²+36x+1>0⇒x-9+0,5√322

D=1296-8=1288

x1 = (-36-2√322) / 4=-9-0,5√322

x2=-9+0,5√322

x∈ (-9+0,5√322; ∞) (основание больше 1)

(2x^2+36x+1) = (x+19) ²

2x²+36x+1-x²-38x-361=0

x²-2x-360=0

x1+x2=2 U x1*x2=-360

x1=20

x2=-18∉ОДЗ

Ответ х=20

Log_x+19 (2x²+36x+1) = log_4 (8) + cos^2 (5pi/4)

log_x+19 (2x²+36x+1) = log_2² (2³) + (-1/√2) ²

log_x+19 (2x²+36x+1) = 3/2+1/2

(х+19) ² = 2 х²+36 х+1

х² + 38 х + 361 = 2 х²+36 х+1

х² - 2 х - 360 = 0

х = 20

х = - 18 не из ОДЗ

ОДЗ

2x²+36x+1 > 0

D/4 = 324-2 = 322 = 2·7·23

x = (-18±√ (322)) / 2

x€ (-∞; - 9-√ (322) / 2) U (-9+√ (322) / 2; + ∞)

-9-√ (322) / 2 ≈ - 17,97

-9+√ (322) / 2 ≈ - 0,028

x+19 > 0

х > - 19

х+19 ≠ 1

х≠ - 18

Ответ 20