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LJ04
@LJ04
July 2022
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2sin^2x-3sinx*cosx+cos^2x=1
решите, пожалуйста
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Answers & Comments
Christmasmole
2sin²x - 3sinx×cosx + cos²x = 1
2sin²x - 3sinx×cosx + cos²x - 1 = 0
2sin²x - 3sinx×cosx + cos²x - (cos²x + sin²x) = 0
2sin²x - 3sinx×cosx + cos²x - cos²x - sin²x = 0
sin²x - 3sinx×cosx = 0
sinx × (sinx - 3cosx) = 0
sinx = 0 или sinx - 3cosx = 0 | ÷(cosx), cosx
x = πn,n⊂Ζ tgx - 3 = 0
tgx = 3
x = arctg3 + πn
1 votes
Thanks 2
Christmasmole
cosx не равное 0
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Answers & Comments
2sin²x - 3sinx×cosx + cos²x - 1 = 0
2sin²x - 3sinx×cosx + cos²x - (cos²x + sin²x) = 0
2sin²x - 3sinx×cosx + cos²x - cos²x - sin²x = 0
sin²x - 3sinx×cosx = 0
sinx × (sinx - 3cosx) = 0
sinx = 0 или sinx - 3cosx = 0 | ÷(cosx), cosx
x = πn,n⊂Ζ tgx - 3 = 0
tgx = 3
x = arctg3 + πn