6. Решить уравнения, сводящиеся к квадратным: a) 2sin²x = 1; 6) 2cos²x + cosx − 3 = 0; B) 3tg²x +tgx − 2 = 0; г) 2cos²x + 3sinx = 0;.

Ответ:

a) 2sin²x = 1

• sin²x = \frac{1}{2}
• sinx = ±\frac{1}{\sqrt{2}} = ±\frac{\sqrt{2}}{2}
• x = ±\frac{\pi}{4} + \pi k, k ∈ Z

б) 2cos²x + cosx - 3 = 0

• Замена: y = cosx
• 2y² + y - 3 = 0
• D = 1² - 4 ⋅ 2 ⋅ (-3) = 1 + 24 = 25
• y₁ = \frac{-1 + 5}{4} = 1
• y₂ = \frac{-1 - 5}{4} = -\frac{3}{2}
• cosx = 1 => x = 2\pi k, k ∈ Z
• cosx = -\frac{3}{2} (нет решений, так как |cosx| ≤ 1)

в) 3tg²x + tgx - 2 = 0

• Замена: y = tgx
• 3y² + y - 2 = 0
• D = 1² - 4 ⋅ 3 ⋅ (-2) = 1 + 24 = 25
• y₁ = \frac{-1 + 5}{6} = \frac{2}{3}
• y₂ = \frac{-1 - 5}{6} = -1
• tgx = \frac{2}{3} => x = arctg(\frac{2}{3}) + \pi k, k ∈ Z
• tgx = -1 => x = -\frac{\pi}{4} + \pi k, k ∈ Z

г) 2cos²x + 3sinx = 0

• 2(1 - sin²x) + 3sinx = 0
• 2 - 2sin²x + 3sinx = 0
• 2sin²x - 3sinx - 2 = 0
• Замена: y = sinx
• 2y² - 3y - 2 = 0
• D = (-3)² - 4 ⋅ 2 ⋅ (-2) = 9 + 16 = 25
• y₁ = \frac{3 + 5}{4} = 2
• y₂ = \frac{3 - 5}{4} = -\frac{1}{2}
• sinx = 2 (нет решений, так как |sinx| ≤ 1)
• sinx = -\frac{1}{2} => x = -\frac{\pi}{6} + 2\pi k, x = \frac{7\pi}{6} + 2\pi k, k ∈ Z

Ответ: a) x = ±\frac{\pi}{4} + \pi k, k ∈ Z; б) x = 2\pi k, k ∈ Z; в) x = arctg(\frac{2}{3}) + \pi k, x = -\frac{\pi}{4} + \pi k, k ∈ Z; г) x = -\frac{\pi}{6} + 2\pi k, x = \frac{7\pi}{6} + 2\pi k, k ∈ Z