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VSKHHolmes41928
@VSKHHolmes41928
July 2022
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[tex]3 sin^{2} 2x+7cos2x-3=0[/tex]
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ribinaolga
3sin²2x + 7cos2x - 3=0
3(1 - cos²2x) + 7cos2x - 3=0
3 - 3cos²2x + 7cos2x -3=0
-3cos²2x + 7cos2x=0
-cos2x(3cos2x - 7)=0
cos2x = 0 3cos2x-7=0
2x=π/2 + πn cos2x=7/3 >1 ОДЗ
x = π/4 + πn/2, n∈Z
Ответ: π/4 + πn/2, n∈Z
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Answers & Comments
3(1 - cos²2x) + 7cos2x - 3=0
3 - 3cos²2x + 7cos2x -3=0
-3cos²2x + 7cos2x=0
-cos2x(3cos2x - 7)=0
cos2x = 0 3cos2x-7=0
2x=π/2 + πn cos2x=7/3 >1 ОДЗ
x = π/4 + πn/2, n∈Z
Ответ: π/4 + πn/2, n∈Z