2*(1-sin^2 (πx/3))+5sin(πx/3)=4; -2sin^2(πx/3)+5sin(πx/3)-2=0
t=sin(πx/3); -2t+5t-2=0; D=25-4*(-2)*(-2)=9=3^2' t1=(-5-3)/(-4)=2;
t2=(-5+3)/(-4)=0,5
sin(πx/3)=0,5 ili sin(πx/3)=2
πx/3=(-1)^n (π/6)+πn решений нет
х=(-1)^n (π/6 * 3/π)+π*(3/π)*n
x=(-1)^n (0,5)+3n, n-celoe
Log₀ ₅ 0.5 log₉ 1/81 - 7^(log₇ 2)= 1 * log₉ 9⁻² - 2 = -2 - 2 = -4
(loga a = 1 a>0 a<>1)
log₃ 6 + log₃ 18 - log₃ 4 = log₃ (6*18 / 4) = log₃ 3³ = 3
9^(0.5 - log₃ 2) - log₃ (log₂ 8)) = (3^2)^0.5 / (3^2)^log₃ 2 - log₃(log₂ 2³)= 3 / 3^log₃ 4 - log₃ 3 = 3/4 -1 = -1/4 =-0.25
1)0,5x²-4x-4=-3
0,5x²-4x-1=0
D=16+2=18 √D=3√2
x1=(4+3√2)/1=(4+3√2) U x2=(4-3√2)
2)0,5x²-4x-4=8
0,5x²-4x-12=0
D=16+24=40 √D=2√10
x1=4+2√10 x2=4-2√10