Question

8cos x=sec x + √3 * csc x

Answer 1

Sec(x) = 1/cos(x); csc(x) = 1/sin(x).
8*cos(x) = ( 1/cos(x) ) + ( (V3)/sin(x) ),
ОДЗ.  cos(x) не=0 и sin(x) не=0.
Делаем замену:
cos(x) = u,
sin(x) = v.
Тогда у нас есть система:
{ 8*u = (1/u) + ( (V3)/v ),
{ u^2 + v^2 = 1.
8*u - (1/u) = ((V3)/v),
v = (V3)/( 8u - (1/u) ) = (V3)*u/( 8*u^2 -1),
1 = u^2 + v^2 = u^2 + ( (V3)*u/( 8*u^2 -1) )^2 = 
 = u^2 + ( 3*u^2/(8*u^2 -1)^2 ),
( 8*u^2 - 1)^2 = u^2*( 8*u^2 -1)^2 + 3*u^2;
u^2 = t,
t = cos^2(x),
0<=t<=1,
(8t -1)^2 = t*(8t-1)^2 + 3t,
64*t^2 - 16t + 1 = t*( 64*t^2 - 16t + 1) + 3t,
64*t^2 -16t + 1 = 64*(t^3) - 16*(t^2) + 4t,
64*(t^3) + t^2*( -16-64) + t*(4+16) - 1 = 0;
64*(t^3) -80*t^2 + 20 t - 1 = 0;
64 = 4^3;
(4t)^3 - 1 - 20t*(4t-1) = 0;
(4t - 1)*( (4t)^2 + 4t + 1 ) - 20t*(4t -1) = 0;
(4t-1)*( 16t^2 + 4t + 1 - 20t) = 0;
(4t -1)*( 16t^2 - 16t +1) = 0;
4t -1 =0 или 16*(t^2) - 16t + 1 = 0;
1) t = 1/4,
cos^2(x) = 1/4,
cos^2(x) = (1+cos(2x))/2 = 1/4,
1+cos(2x) = 1/2,
cos(2x) = -1/2,
....
2) 16*(t^2) - 16t + 1 = 0;
D/4 = 8^2 - 16 = 64 - 16 = 48 = 16*3 = 3*(4^2).
t1 = (8 - 4*(V3) )/16 = (2 - V3)/4;
t2 = (8+4*(V3))/ 16 = (2 + V3)/4.
....
дорешайте сами, я устал печатать.