Question

7 - 6cos^2=5 sin x уравнение
6 sin^2 x - 5sin x + 1=0 приведи это к этой форме
помогите!!! СРОЧНО
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Answer 1

7 - 6cos^2 x = 5 sin x  // cos²x + sin²x = 1 => cos²x = 1 - sin²x

7 - 6( 1 - sin²x) = 5sinx

7 - 6 + 6sin²x - 5sinx = 0

6 sin²x - 5sinx + 1 = 0

sinx = t, sin²x = t²

6t² - 5t +1 =0

D =b²-4ac = (-5)²-4*6*1 = 25-24 =1

$t1 = \frac{-b+\sqrt{D} }{2a} = \frac{5+1}{2*6} = \frac{6}{12} =\frac{1}{2} \\\\t2 = \frac{-b-\sqrt{D} }{2a} = \frac{5-1}{2*6} = \frac{4}{12} =\frac{1}{3}$

sin x = 1/2

x = (-1)^n * π/6 +πn

sin x = 1/3

x = (-1)^n * arcsin(1/3) + πn