Question

8sin^2(x/2)-3sinx-4=0

Answer 1

8sin²(x/2)-6sin(x/2)cos(x/2)-4cos²(x/2) - 4sin²(x/2)=0

4sin²(x/2)-6sin(x/2)cos(x/2)-4cos²(x/2)=0|impartim cu 2

2sin²(x/2)-3sin(x/2)cos(x/2)-2cos²(x/2)

impartim cu cos²(x/2)≠0,deci punem conditia x/2≠(2k+1)π/2, x≠(2k+1)π

2tg²(x/2)-3tg(x/2)-2=0

tg(x/2) =t

2t²-3t-2=0

t1,2= (3+/-√(9+16))/4= (3+/-5)/4

t1=-1/2=tg(x1/2)⇒....x1/2 =kπ+arctg(-1/2)...x1=2kπ+2arctg(-1/2), k∈Z

t2=2=tg(x2/2)⇒...x2/2=kπ+arctg2.......x2=2kπ+2arctg2, K∈Z