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araykooosh
@araykooosh
June 2022
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Rokcacke
А) sin(Pi/4-x)=1/2
1)Pi/4 - x = Pi/6+2Pi*n
-x=Pi/6-Pi/4+2Pi*n |*(-1)
x=Pi/12+2Pi*n
2)Pi/4 - x = 5Pi/6+2Pi*n
-x=5Pi/6-Pi/4+2Pi*n |*(-1)
x= -7Pi/12 + 2Pi*n
Oтвет: Pi/12+2Pi*n ; -7Pi/12 + 2Pi*n
б)√3sinx*cosx=sin^2x
√3sinx*cosx-sin^2x=0
sinx(√3cosx-sinx)=0
sinx=0. √3cosx-sinx=0 |:cosx
x=Pi*n. √3-tgx=0
tgx=√3
x=Pi/3+2Pi*n
Ответ:Pi*n; Pi/3+2Pi*n
Не забудьте везде написать , что n принадлежит z
1 votes
Thanks 1
araykooosh
огромное спасибо за ответ
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Answers & Comments
1)Pi/4 - x = Pi/6+2Pi*n
-x=Pi/6-Pi/4+2Pi*n |*(-1)
x=Pi/12+2Pi*n
2)Pi/4 - x = 5Pi/6+2Pi*n
-x=5Pi/6-Pi/4+2Pi*n |*(-1)
x= -7Pi/12 + 2Pi*n
Oтвет: Pi/12+2Pi*n ; -7Pi/12 + 2Pi*n
б)√3sinx*cosx=sin^2x
√3sinx*cosx-sin^2x=0
sinx(√3cosx-sinx)=0
sinx=0. √3cosx-sinx=0 |:cosx
x=Pi*n. √3-tgx=0
tgx=√3
x=Pi/3+2Pi*n
Ответ:Pi*n; Pi/3+2Pi*n
Не забудьте везде написать , что n принадлежит z