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July 2022
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Как решить такое уравнение? 3cos^3x-4cosx+1=0
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yulyanataran
T=cos x
3t3-4t+1=0
a= 3•3=9 b=-4 c= 1
D= b2 - 4ac= 16+4•9•1=√52
√D = √52
t1= -b+(-)√D. 4+52. 56. 28
------------ = ----------- = --------- = -------
2a. 2•9. 18. 9
4-52. -48. -16. - 8
t2= ---------- = ------- = -------- = -----
18. 18. 6. 3
28. 28
[ cos x = ----- [ x = +(-) arccos ----- + 2πn;
9. => 9
-8. -8
[ cos x = ------ [ x = +(-) arccos ----- + 2πn;
3. 3
как-то так. возможно и неверно
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Answers & Comments
3t3-4t+1=0
a= 3•3=9 b=-4 c= 1
D= b2 - 4ac= 16+4•9•1=√52
√D = √52
t1= -b+(-)√D. 4+52. 56. 28
------------ = ----------- = --------- = -------
2a. 2•9. 18. 9
4-52. -48. -16. - 8
t2= ---------- = ------- = -------- = -----
18. 18. 6. 3
28. 28
[ cos x = ----- [ x = +(-) arccos ----- + 2πn;
9. => 9
-8. -8
[ cos x = ------ [ x = +(-) arccos ----- + 2πn;
3. 3
как-то так. возможно и неверно