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artemy050702
@artemy050702
July 2022
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Решите тригонометрическое уравнение
3sin5z-2cos5z=3
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армения20171
3sin5z-2cos5z=3
√13(3/√13sin5z-2/√13cos5z)=3
cosa=3/√13;sina=2/√13
sin(5z-a)=3/√13
5z-a=(-1)ⁿarcsin3/√13+πn
z=(arccos3√13+(-1)ⁿarcsin3/√13+πn)*1/5
n=2k;z=
(arccos3/√13+arcsin3/√13
+π*2k)*1/5=
(π/2+2πk)*1/5=π/10+2πk/5
n=2k+1
z=(arccos3/√13-arcsin3/√13+2πk+π)*1/5=
(π/2-2arcsin3/√13+2πk+π)*
1/5=3π/10-2/5arcsin3/√13
+2πk/5
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Answers & Comments
√13(3/√13sin5z-2/√13cos5z)=3
cosa=3/√13;sina=2/√13
sin(5z-a)=3/√13
5z-a=(-1)ⁿarcsin3/√13+πn
z=(arccos3√13+(-1)ⁿarcsin3/√13+πn)*1/5
n=2k;z=
(arccos3/√13+arcsin3/√13
+π*2k)*1/5=
(π/2+2πk)*1/5=π/10+2πk/5
n=2k+1
z=(arccos3/√13-arcsin3/√13+2πk+π)*1/5=
(π/2-2arcsin3/√13+2πk+π)*
1/5=3π/10-2/5arcsin3/√13
+2πk/5