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

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

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

Решений нет

$x=\pm\frac{\pi }{4}+\pi n,n\in z$

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