Решение
<span>5sinx+cosx=5
Применяя формулы:
sinx = sin2*(x/2); cosx = cos2*(x/2)
sin</span>²x/2 + cos²x/2 = 1
Получим уравнение:
5* sin2*(x/2) + cos2*(x/2) = 5*(sin<span>²x/2 + cos²x/2)
5*(2sinx/2 * cosx/2) + (cos</span>²x/2 - sin²x/2) = 5*(sin<span>²x/2 + cos²x/2)
10</span>sinx/2 * cosx/2 + cos²x/2 - sin²x/2 - 5sin<span>²x/2 - 5cos²x/2 = 0
- 6sin</span>²x/2 + 10sinx/2 * cosx/2 - 4cos²x/2 = 0 делим на (- 2cos²x/2 ≠ 0)
3tg²x/2 - 5tgx + 2 = 0
tgx = t
3t² - 5t + 2 = 0
D = 25 - 4*3*2 = 1
t₁ = (5 - 1)/6 = 4/6 = 2/3
t₂ = (5 + 1)/6 = 6/6 = 1
tgx = 2/3
x₁ = arctg(2/3) + πk, k ∈ Z
tgx = 1
x₂ = π/4 + πn, n ∈ Z
1) log₀.₂₅ (2x²-7x-6)= -2
ОДЗ: 2x²-7x-6>0
2x²-7x-6=0
D=49+48=97
x₁= <u>7-√97</u> ≈ -0.71
4
x₂ = <u>7+√97 </u>≈ 4.21
4
+ - +
------------ -0.71 ------------ 4.21 -------------
\\\\\\\\\\\\\\\ \\\\\\\\\\\\\\\
x∈(-∞; -0,71)U(4,21; +∞)
log₀.₂₅ (2x²-7x-6)=log₀.25 (0.25)⁻²
2x²-7x-6 =0.25⁻²
2x²-7x-6=(1/4)⁻²
2x²-7x-6=4²
2x²-7x-6-16=0
2x²-7x-22=0
D=49-4*2(-22)=49+176=225
x₁= <u>7 -15 </u>= -8/4= -2
4
x₂=<u> 7+15</u> = 22/4 = 5.5
4
Ответ: -2; 5,5
2) log₀.₅ (x-4)<1
ОДЗ: х-4>0
x> -4
log₀.₅ (x-4) < log₀.5 0.5
x-4>0.5
x>0.5+4
x>4.5
3) log₂ x +log₄ x + log₁₆ x > 3.5
log₂ x +log₂² x +log₂⁴ x >3.5
log₂ x +log₂ x^(¹/₂) +log₂ x^(¹/₄) > 3.5
log₂ (x*x^(¹/₂)*x^(¹/₄)) > log₂ 2^(3.5)
log₂ (x^(⁷/₄)) > log₂ 2^(⁷/₂)
x^(⁷/₄) > 2^(⁷/₂)
(x^(¹/₂))^(⁷/₂) > 2^(⁷/₂)
√x >2
x>4
раскрываем скобки,используя формулу.ничего сложного
-50tg9*tg(90-9)+31= -50tg9*ctg9+31= -50+31= -19