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yaaaaaaan
@yaaaaaaan
July 2022
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22cos^2x+8sinx*cosx=7
Помогите решить простейшее тригонометрическое уравнение, если можно, то напишите на листочке решение, а ответ сюда прислать.
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ShirokovP
Verified answer
22cos^2x + 8sinx*cosx = 7
22cos^2x + 8sinx*cosx = 7(cos^2x + sin^2x)
22cos^2x + 8sinx*cosx = 7cos^2x + 7sin^2x)
- 7sin^2x + 8sinxcosx + 15cos^2x = 0 /:cos^2x ≠ 0
- 7tg^2x + 8tgx + 15 = 0
7tg^2x - 8tgx - 15 = 0
tgx = t
7t^2 - 8t - 15 = 0
D = 64 + 4*15*7 = 484 = 22^2
t1 = ( 8 + 22)/14 = 30/14 = 15/7
t2 = ( 8 - 22)/14 = - 14/14 = - 1
1) tgx = 15/7
x = arctg(15/7) + pik, k ∈ Z
2) tgx = - 1
x = - pi/4 + pik, k ∈ Z
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Answers & Comments
Verified answer
22cos^2x + 8sinx*cosx = 722cos^2x + 8sinx*cosx = 7(cos^2x + sin^2x)
22cos^2x + 8sinx*cosx = 7cos^2x + 7sin^2x)
- 7sin^2x + 8sinxcosx + 15cos^2x = 0 /:cos^2x ≠ 0
- 7tg^2x + 8tgx + 15 = 0
7tg^2x - 8tgx - 15 = 0
tgx = t
7t^2 - 8t - 15 = 0
D = 64 + 4*15*7 = 484 = 22^2
t1 = ( 8 + 22)/14 = 30/14 = 15/7
t2 = ( 8 - 22)/14 = - 14/14 = - 1
1) tgx = 15/7
x = arctg(15/7) + pik, k ∈ Z
2) tgx = - 1
x = - pi/4 + pik, k ∈ Z