Question

1) 12sin5x=cos10x+7 2) 3tg²x -8cos²x+1=0 3) 3sinx+cosx=1

Answer 1

1) 12Sin5x = Cos10x + 7

12Sin5x - Cos10x - 7 = 0

12Sin5x - (1 - 2Sin²5x) - 7 = 0

12Sin5x - 1 + 2Sin²5x - 7 = 0

2Sin²5x + 12Sin5x - 8 = 0

Sin²5x + 6Sin5x - 4 = 0

Sin5x = m ,  - 1 ≤ m ≤ 1

m² + 6m - 4 = 0

D = 6² - 4 * (- 4) = 36 + 16 = 52 = (2√13)²

$m_{1}=frac{-6-2sqrt{13}}{2}=-(3+sqrt{13})<-1\\m_{2}=frac{-6+2sqrt{13} }{2}=sqrt{13}-3\\Sin5x=sqrt{13}-3\\5x=(-1)^{n}arcSin(sqrt{13}-3)+pi n,nin z\\x=(-1)^{n}frac{1}{5}arcSin(sqrt{13}-3)+frac{pi n }{5},nin z$

$2)3tg^{2}x-8Cos^{2}x+1=0\\3tg^{2}x-8*frac{1}{1+tg^{2}x }+1=0\\3tg^{2}x+3tg^{4}x-8+1+tg^{2}x=0\\3tg^{4}x+4tg^{2}x-7=0\\1)tg^{2}x=1\\x=pmfrac{pi }{4}+pi n,nin z\\2)tg^{2}x=-frac{7}{3}$

Решений нет

Ответ:

$x=pmfrac{pi }{4}+pi n,nin z$

$3)3Sinx+Cosx=1\\ASinx+BCosx=CSin(x+t)\\C=sqrt{A^{2}+B^{2}}=sqrt{3^{2}+1^{2}}=sqrt{10}\\sqrt{10}Sin(x+t)=1,t=arcSinfrac{1}{sqrt{10} } \\x+t=(-1)^{n} frac{1}{sqrt{10} }+pi n,nin z\\x=(-1)^{n}arcSinfrac{1}{sqrt{10} }-arcSinfrac{1}{sqrt{10} }+pi n,nin z$