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gogen228336
@gogen228336
July 2022
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Решить уравнение:
sin2x=cos2x-sin^2x+1
( sin^2x - синус квадрат x)
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Answers & Comments
армения20171
Sin2x=cos2x-sin²x+1
2sinxcosx=cos²x-sin²x-sin²x+(sin²x+cos²x)
2sinxcosx=2cos²x-sin²x ; :cos²x≠0
2tgx=2-tg²x
tg²x+2tgx-2=0
tgx=t
t²+2t-2=0
D=4+8=12=4*3
t=(-2±2√3)/2=(-1±√3)
t1=-(1+√3);t2=-1+√3
1)tgx=-(1+√3)
x=-arctg(1+√3)+πk;k€Z
2)tgx=(√3-1)
x=arctg(√3-1)+πk;k€Z
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Answers & Comments
2sinxcosx=cos²x-sin²x-sin²x+(sin²x+cos²x)
2sinxcosx=2cos²x-sin²x ; :cos²x≠0
2tgx=2-tg²x
tg²x+2tgx-2=0
tgx=t
t²+2t-2=0
D=4+8=12=4*3
t=(-2±2√3)/2=(-1±√3)
t1=-(1+√3);t2=-1+√3
1)tgx=-(1+√3)
x=-arctg(1+√3)+πk;k€Z
2)tgx=(√3-1)
x=arctg(√3-1)+πk;k€Z