Решите уравнение f'(x)=0
f(x)=3cosx+4sinx-5x
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Ответил ShirokovP
0
f'(x)= - 3sinx + 4cosx - 5
4 cosx - 3sinx = 5
16cos^2x - 24sinxcosx + 9sin^2x = 25*1
16cos^2x - 24sinxcosx + 9sin^2x = 25*(sin^2x + cos^2x)
16cos^2x - 24sinxcosx + 9sin^2x = 25sin^2x + 25cos^2x
- 9cos^2x - 24sinxcosx - 16 sin^2x = 0 /:(-1)
9cos^2x + 24sinxcosx + 16sin^2x = 0
(3cosx + 4sinx)^2 = 0
3cosx + 4sinx = 0 /:cosx ≠ 0
4tgx + 3 = 0
4tgx = - 3
tgx= - 3/4
x= - arctg(3/4) + pik, k ∈Z
4 cosx - 3sinx = 5
16cos^2x - 24sinxcosx + 9sin^2x = 25*1
16cos^2x - 24sinxcosx + 9sin^2x = 25*(sin^2x + cos^2x)
16cos^2x - 24sinxcosx + 9sin^2x = 25sin^2x + 25cos^2x
- 9cos^2x - 24sinxcosx - 16 sin^2x = 0 /:(-1)
9cos^2x + 24sinxcosx + 16sin^2x = 0
(3cosx + 4sinx)^2 = 0
3cosx + 4sinx = 0 /:cosx ≠ 0
4tgx + 3 = 0
4tgx = - 3
tgx= - 3/4
x= - arctg(3/4) + pik, k ∈Z
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