1) sin x = √2/2
x = (-1)ⁿ × arcsin √2/2 + πn, n∈Z
x = (-1)ⁿ × π/4 + πn, n∈Z
2) sin x = -√2/2
x = (-1)ⁿ × arcsin (-√2/2) + πn, n∈Z
x = (-1)ⁿ × -arcsin √2/2 + πn, n∈Z
x = (-1)ⁿ × (-π/4) + πn, n∈Z
3) sin x = -√3/2
x = (-1)ⁿ × arcsin (-√3/2) + πn, n∈Z
x = (-1)ⁿ × -arcsin √3/2 + πn, n∈Z
x = (-1)ⁿ × (-π/3) + πn, n∈Z
4) sin x = √3/2
x = (-1)ⁿ × arcsin √3/2 + πn, n∈Z
x = (-1)ⁿ × arcsin √3/2 + πn, n∈Z
x = (-1)ⁿ × π/3 + πn, n∈Z
5) sin x = 4/5
x = (-1)ⁿ × arcsin 4/5 + πn, n∈Z
x = (-1)ⁿ × 0,927295 + πn, n∈Z
x = (-1)ⁿ × 53,1° + πn, n∈Z
9.
log₁₄ 7 = m найдем log₁₇₅ 56 - ?
log₁₄ 5 = n
Используем формулу перехода другому основанию:
log₁₇₅ 56 = log₁₄ 56/log₁₄ 175 = log₁₄ (8×7)/log₁₄ (25×7) = log₁₄ (2³×7)/log₁₄ (5²×7) = log₁₄ 2³ × log₁₄ 7/log₁₄ 5² × log₁₄ 7 = 3log₁₄ 2 × log₁₄ 7/2log₁₄ 5 × log₁₄ 7
Нам нужно найти log₁₄ 2:
log₁₄ 2 = log₁₄ 14/7 = log₁₄ 14 - log₁₄ 7 = 1 - m
Получаем:
log₁₇₅ 56 = 3×(1 - m) + m/2n + m = 3 - 3m + m/2n + m = 3 - 2m/2n + m
ответ: log₁₇₅ 56 = 3 - 2m/2n + m
10.
log₅ 5 = 1
log₁₁ 15 = log₁₀ 15/log₁₀ 11 ≈ 1,17609/1,04139 ≈ 1,12934
Следовательно:
1 < 1,12934
log₅ 5 < log₁₁ 15
ответ: log₅ 5 < log₁₁ 15
