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Ответ дан
kirichekov

Ответ:

x_{1}=-arctgfrac{1}{2}+pi n,n∈Z

x_{2}=arctgfrac{1}{3}+pi n,n∈Z

Объяснение:

5sin²x+sinx*cosx=2cos²x-1

5sin²x+sinx*cosx=2cos²x-(sin²x+cos²x)

6sin²x+sinx*cosx-cos²x=0 | : cos²x≠0

frac{6sin^{2}x}{cos^{2}x}+frac{sinx*cosx}{cos^{2}x}-frac{cos^{2}x}{cos^{2}x}=0\  6*(frac{sinx}{cosx})^{2}+frac{sinx}{cosx} -1=0

6tg²x+tgx-1=0 тригонометрическое квадратное уравнение, замена переменной:

tgx=t

6t²+t-1=0. t₁=-1/2, t₂=1/3

обратная замена:

t₁=-1/2,  tgx=-1/2,  x=arctg(-frac{1}{2})+pi n,n∈Z

x=-arctgfrac{1}{2}+pi n,n∈Z

t₂=1/3, tgx=1/3, x=arctgfrac{1}{3}+pi n,n∈Z  

Ответ

Ответ дан
sobakazabiyaka

Ответ в закрепе..........

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