Question

$2sin^2x+3\sqrt{2} sin(3\pi /2-x)-4=0$ на отрезке [pi ; 5pi/2]

Answer 1

Ответ:

Объяснение:

2sin²x - 3√2cosx - 4 = 0

2(1 - cos²x ) - 3√2cosx - 4 = 0

2 - 2cos²x - 3√2cosx - 4 = 0

-2cos²x - 3√2cosx - 2 = 0

2cos²x + 3√2cosx + 2 = 0

Замена: cosx = t, t ∈ [-1 ; 1]

2t² + 3√2t + 2 = 0

D = 18 - 16 = 2

t₁ = $\frac{-3\sqrt{2} + \sqrt{2} }{4} =-\frac{\sqrt{2} }{2}$

t₂ = $\frac{-3\sqrt{2} - \sqrt{2} }{4} =-\frac{\sqrt{2} }{2}$ ∉ [-1 ; 1]

cosx = $\frac{\sqrt{2} }{2}$

x = ± $\frac{3\pi }{4}$

На отрезке

x = $\frac{5\pi }{4}$