8cos x=sec x + √3 * csc x

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.