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ksharifi
@ksharifi
July 2022
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1)Решить уравнение:
4sin2x-3cos2x=3
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Nella4ka
4*2sinx*cosx-3(cosx^2-sinx^2)=3
8sinx*cosx-3cosx^2+3sinx^2=3
8sinx*cosx-3cosx^2+3(1-cosx^2)=3
8sinx*cosx-6cosx^2+3=3
8sinx*cosx-6cosx^2=0
2cosx*(4sinx-3cosx)=0
Cosx=0 x=pi/2 +pin
Или 4sinx-3cosx=0 x=arctg 3/4 +pin
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Answers & Comments
8sinx*cosx-3cosx^2+3sinx^2=3
8sinx*cosx-3cosx^2+3(1-cosx^2)=3
8sinx*cosx-6cosx^2+3=3
8sinx*cosx-6cosx^2=0
2cosx*(4sinx-3cosx)=0
Cosx=0 x=pi/2 +pin
Или 4sinx-3cosx=0 x=arctg 3/4 +pin