Question

$2sin^2 x + 6sin 2x = 7(1 + cos 2x)$

Answer 1

2sin^2x + 6*2sinxcosx = 7 + 7cos2x 
2sin^2x + 12sinxcosx = 7cos^2x + 7sin^2x + 7cos^2x - 7sin^2x
2sin^2x + 12sinxcosx = 14cos^2x 
2sin^2x + 12sinxcosx - 14cos^2x  = 0  /:2
sin^2x + 6sinxcosx - 7cos^2x  = 0  /:cos^2x ≠ 0
tg^2x + 6tgx - 7 = 0
Пусть tgx = t, тогда
t^2 + 6t - 7 = 0 
D = 36 + 28 = 64 = 8^2
t1 = ( - 6 + 8)/2 = 2/2 = 1;
t2 = ( - 6 - 8)/2 = - 14/2 = - 7

1) tgx = 1 
x = pi/4 + pik, k ∈Z

2) tgx = - 7
x = - arctg7 + pik, k ∈Z