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Б45
@Б45
July 2022
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6sin^2x+sin2x=4cos^2x.Решите пожалуйстауравнение,ответ должен быть -п/4+пn
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Christmasmole
6sin²x + sin2x = 4cos²x
6sin²x + 2sinxcosx - 4cos²x = 0 | ÷(cos²x),где cosx≠0
6tg²x + 2tgx - 4 = 0 |÷2
3tg²x + tgx - 2 = 0
D = 5²
tgx = -1
x = -
+πn
tgx =
x = arctg
+ πn
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Answers & Comments
6sin²x + 2sinxcosx - 4cos²x = 0 | ÷(cos²x),где cosx≠0
6tg²x + 2tgx - 4 = 0 |÷2
3tg²x + tgx - 2 = 0
D = 5²
tgx = -1
x = - +πn
tgx =
x = arctg + πn