Ну я думаю.что ответом будет число 16,5
∠с=180-2*53=74
∠акс=180-53-74/2=90
1
cosa=-√(1-sin²a)=-√(1-0,36)=-√0,64=-0,8
2
(tg(7π/16)-tg(3π/16))/(1+tg(7π/16)*tg(3π/16))=tg(7π/16-3π/16)=tgπ/4=1
3
1/(2sin2acosa)+1/(2sin4acosa)=(sin4a+sin2a)/(2sin2acosasin4a)=
=2sin3acosa/<span>(2sin2acosasin4a)=sin3a/(sin4asin2a)
</span>a=π/12
sin(π/4):(sin(π/6)sin(π/3)=√2/2:(1/2*√3/2)=√2/2:√3/4=√2/2*4/√3=
=2√2/√3=2√6/3
7) log₀,₅(x + 8) - log₀,₅(x - 3) > log₀,₅3x;
ОДЗ: x > -8; Имеем: x > 3.
x > 3;
x > 0
log₀,₅(x + 8) > log₀,₅3x + log₀,₅(x - 3);
log₀,₅(x + 8) > log₀,₅3x(x - 3);
x + 8 < 3x(x - 3);
3x² - 9x - x - 8 > 0;
3x² - 10x - 8 > 0;
3x² - 10x - 8 = 0; D = 100 + 96 = 196; √D = 14;
x₁ = (10 + 14)/6 = 4; x₂ = (10 - 14)/6 = -4/6 = -2/3
------ ++++
---------------------3----------------4----------->
x∈(4; ∞).
Ответ: (4; ∞).
8) log²₃(27x) + log₃(x³/9) = 17;
ОДЗ: x > 0
(log₃27 + log₃x)² + log₃(x³) - log₃9 = 17;
(3 + log₃x)² + 3log₃x - 2 - 17 = 0;
9 + 6log₃x + log²₃x + 3log₃x - 19 = 0;
log²₃x + 9log₃x - 10 = 0. Замена: log₃x = t
t² + 9t - 10 = 0;
t₁ = -10; t₂ = 1.
Обратная замена:
log₃x = -10 или log₃x = 1
x₁ = 3⁻¹⁰ x₂ = 3
Ответ: 3⁻¹⁰; 3.