1wangshu 1wangshu
  • 10-10-2020
  • Mathematics
contestada

Calculate the number of roots of
2sin^3x-5sin^2x+2sinx=0
in the interval 0<=x<=pi

Respuesta :

Аноним Аноним
  • 10-10-2020

Answer:

Step-by-step explanation:

2sin³x - 5sin²x + 2sinx = 0(*)

suppose : sinx = t; t∈[-1;1]

(*)⇔ 2t³ - 5t² +2t = 0

⇔ t.(2t² - 5t +2) = 0

⇔ t( t -2)(2t -1) = 0

⇔ t = 0

    t = 2 (unsatisfactory)

     t = 1/2

⇔ sin x = 1

     sin x = 1/2

⇔ x = pi

    x = pi/6

( because 0≤x≤pi)

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