Помогите, пожалуйста, кто чем может)
3cos²x+11sinx+6=0
7cos²x-5sinx-5=0
5tgx+4ctgx+21=0
cosx-6sinx=0
5sin²x+4sinx=0
sin6x+sin2x=0
5sin2x-6cosx=0
3sin2x+4cos²x=0
4cos2x+44sin²x-29=0
4cos2x-10cosx+1=0
5sin2x+5sin²x+3=0 1
1cosx-7sinx+1=0
cosx-6sinx=0 |разделим на cosx≠0
1-6tgx=0
tgx=1/6
x=arctg1/6+πn, n∈Z
5sin2x-6cosx=0
10sinxcosx-6cosx=0
2cosx(5sinx-3)=0
cosx=0 или 5sinx-3=0
x=π/2+πn, n∈Z 5sinx=3
sinx=3/5
x=(-1)^n*arcsin(3/5)+2πn, n∈Z
7cos²x-5sinx-5=0
7(1-sin²x)-5sinx-5=0
7-7sin²x-5sinx-5=0
7sin²x+5sinx-2=0
введем замену переменной sinx=t
7t²+5t-2=0
D=25+56=81
t₁=(-5+9)/14=2/7
t₂=(-5-9)/14=-1
вернемся к замене
sinx=2/7
x=(-1)^n*arcsin(2/7)+2πn, n∈Z
sinx=-1
x=-π/2+2πn, n∈Z
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