Помогите пожалуйста срочно надо
3. 6(1 - cos²x) - cosx - 5 =0; 6 - 6cos²x - cosx - 5 = 0; 6cos²x - cosx -1 = 0 cosx = y; 6y² + y - 1 = 0; y= - 1/2; y = 1/3; cosx = +- 2π/3 +2πn, n∈Z. cosx = arccos1/3 +2πn; n∈z. 1)
![\sqrt{25+3 -2*5 \sqrt{3} } + \sqrt{25+3-2*5 \sqrt{3} } = [tex] \sqrt{ (5 + \sqrt{3} )^{2} } + \sqrt{ (5- \sqrt{x3 )^{2} } = 10;](https://tex.z-dn.net/?f=+%5Csqrt%7B25%2B3+-2%2A5+%5Csqrt%7B3%7D+%7D++%2B++%5Csqrt%7B25%2B3-2%2A5+%5Csqrt%7B3%7D+%7D+%3D+%C2%A0%5Btex%5D+%5Csqrt%7B+%285+%2B+%5Csqrt%7B3%7D+%29%5E%7B2%7D+%7D+%2B+%5Csqrt%7B+%285-+%5Csqrt%7Bx3+%29%5E%7B2%7D+%7D+%3D++10%3B "\sqrt{25+3 -2*5 \sqrt{3} } + \sqrt{25+3-2*5 \sqrt{3} } = [tex] \sqrt{ (5 + \sqrt{3} )^{2} } + \sqrt{ (5- \sqrt{x3 )^{2} } = 10;")