1. 1) √2cosx - 1 = 0,
√2cosx = 1,
cosx = 1/√2,
cosx = √2/2,
x = +-π/4 + 2πn, n ∈ Z.
ответ: +-π/4 + 2πn, n ∈ Z.
2) 3tg2x + √3 = 0,
3tg2x = - √3,
tg2x = -√3/3,
2x = -π/6 + πn, n ∈ Z,
x = -π/12 + πn/2, n ∈ Z.
ответ: -π/12 + πn/2, n ∈ Z.
2. sinx/3 = -1/2 на [0; 3π]
x/3 = -π/6 · (-1)ⁿ + πn, n ∈ Z,
x = π/2 · (-1)ⁿ⁺¹ + 3πn, n ∈ Z.
Найдем корни из [0; 3π]:
при n = 0 x = π/2 · (-1) = -π/2 ∉ [0; 3π],
при n = 1 x = π/2 · (-1)² + 3π = π/2 + 3π ∉ [0; 3π],
при n = -1 x = π/2 · (-1)⁰ - 3π = -5π/2 ∉ [0; 3π],
нет решений на [0; 3π].
ответ: нет решений на [0; 3π].
3. 1) 3cosx - cos²x = 0,
cosx(3 - cosx) = 0,
cosx = 0 (1) или 3 - cosx = 0 (2),
(1): x = π/2 + πn, n ∈ Z;
(2): 3 - cosx = 0,
cosx = 3 - нет решений.
Овет: π/2 + πn, n ∈ Z.
2) 6sin²x - sinx = 1,
sinx = t
6t² - t - 1 = 0,
D = (-1)² - 4 · 6 · (-1) = 1 + 24 = 25; √25 = 5
t₁ = (1 + 5)/(2 · 6) = 6/12 = 1/2,
t₂ = (1 - 5)/(2 · 6) = -4/12 = -1/3.
sinx = 1/2,
x = π/6 · (-1)^m + πm, m ∈ Z;
sinx = -1/3,
x = (-1)ⁿ · arcsin(-1/3) + πn, n ∈ Z,
x = (-1)ⁿ⁺¹· arcsin(1/3) + πn, n ∈ Z.
ответ: π/6 · (-1)^m + πm, m ∈ Z; (-1)ⁿ⁺¹· arcsin(1/3) + πn, n ∈ Z.
Начинаем решать:
y=(2•(-2)-5):4=-1,75
y=(2•0):4=0
y=(2•1):4=0,5.