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xzloyx
@xzloyx
October 2021
1
9
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решите неравенство
2cos ( 2x + пи/3) меньше или равно 1
с единичной окружностью
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Anytim
1) 5sin^2x+ 6 cos x - 6= 0
5(1-cos^2x) + 6cosx - 6 = 0
5 - 5cos^2x + 6cosx - 6 = 0
-5cos^2x + 6cosx - 1 = 0
пусть cosx - y тогда
-5y^2 + 6y - 1 = 0
D = 36 - 4 *(-5)*(-1) = 16
y= 1 ; y= 1/5 значит
cosx = 1 cosx = 1/5
x = 2пk, x = +-arccos 1/5 + 2пk
Ответ: +-arccos 1/5 + 2пk; 2пk.
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Answers & Comments
5(1-cos^2x) + 6cosx - 6 = 0
5 - 5cos^2x + 6cosx - 6 = 0
-5cos^2x + 6cosx - 1 = 0
пусть cosx - y тогда
-5y^2 + 6y - 1 = 0
D = 36 - 4 *(-5)*(-1) = 16
y= 1 ; y= 1/5 значит
cosx = 1 cosx = 1/5
x = 2пk, x = +-arccos 1/5 + 2пk
Ответ: +-arccos 1/5 + 2пk; 2пk.