\\\\\\\\\\\\\\\\\\\\\\\\\\
1) - 3х^2+4х-7=0 - корней нет
![2) \: 2 {x}^{2} - 7x + 8 = 0 \\ \\ d = ( - 7)^{2} - 4 \times 2 \times 8 = 49 - 64 = - 15](https://tex.z-dn.net/?f=2%29+%5C%3A+2+%7Bx%7D%5E%7B2%7D++-+7x+%2B+8+%3D+0+%5C%5C++%5C%5C+d+%3D+%28+-+7%29%5E%7B2%7D++-+4+%5Ctimes+2+%5Ctimes+8+%3D+49+-+64+%3D++-+15+)
корней нет
<span>task/26087093
--------------------- </span> аккуратно с аккуратно с заданием <span>, </span><span>пожалуйста
</span><span>cos^6 x +sin^6 x - cos^2 x = 1/16 ;
(cos</span>²x)³ +(sin²x)³ - cos²x =1/16 ; * * *a³+b³=(a+b)³ - 3ab(a+b) * * *
(cos²x +sin²x)³ -3sin²x*cos²x(cos²x +sin²x) - cos²x = 1/16 ;
1 - 3sin²x*cos²x - cos²x = 1/16 * * * если 1 - 3/4sin²2x - cos²2x =1/16 * * *
1 -3(1-cos²x)cos²x -cos²x -1/16=0 ;
3cos<span>⁴x - 4cos</span>²x +15/16 =0 ;
48cos⁴x - 64cos²x +15 =0 ;
* * * биквадратное уравнение * * * t =cos²x , 0 ≤ t ≤1
48t² - 64t +15 =0 ; D₁ = 32² -48*15= 16 *19 =(4√19)²<span>
t</span>₁= (32+4√19) /48=(8 +√19)/12 > (8+<span>√16) /12 =1</span> не удовлетворяет<span>;
</span>t₂ = (8-√19) /12 .
cos²x = (8-<span>√19) /12 ;
(1+cos2x )/2 = </span>(8-<span>√19) /12 ;
cos2x = </span> -(√19 -2<span>) /6
2x = </span>± (π -arccos( (√19 -2) /6 ) +2πn , n∈Z;
x = ±0,5 (π -arccos( (√19 -2) /6 )<span> +πn , n∈Z.
</span>
ответ: x = ±0,5(π -arccos( (√19 -2) /6 ) +πn , n∈Z .
* * * * * * * * * * * * * * * * * * * * * * * * *
А Если ...
<span>cos^6 x +sin^6 x - cos^2 2x = 1/16;
--------------
</span>1 - 3sin²x*cos²x - cos²2x = 1/16 ;
1 -(3/4)sin²2x - cos²2x =1/16 ;
sin²2x - (3/4)sin²<span>2x =1/16 ;
sin</span>²<span>2x = 1/4 ;
(1-cos4x) /2 =1/4 ;
cos4x = 1/2 ;
4x = </span>±π/3 +2πn , n∈Z ;
x = ±π/12 +(π/2)*n , n∈Z.
-------
Удачи !