Question

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5sin^2x-sinxcosx=2

Answer 1

5sin²x - sinx·cosx = 2(sin²x + cos²x)
5sin²x - sinx·cosx - 2sin²x - 2cos²x = 0
3sin²x - sinx·cosx - 2cos²x = 0    однородное, делим на cos²x:
3tg²x - tgx - 2 = 0
D = 1 + 24 =25
tgx = (1 + 5)/6 = 1    или    tgx = (1 - 5)/6 = - 2/3
x = π/4 + πn                      x =  - arctg(2/3)  + πk