4cos2x=2cos(π/2-x)+1
4(1-2sin²x)=2sinx+1
4-8sin²x-2sinx-1=0,Замена sinx=t,t[-1;1]
-8t²-2t+3=0 |•(-1)
8t²+2t-3=0
D=4+3*4*8=96+4=100=10²
t1=(-2-10)/16=-12/16=-3/4
t2=(-2+10)/16=8/16=1/2
1)sinx=1/2
x1=π/6+2πk,k-Z x2=5π/6+2πn,n-Z
2)sinx=-3/4
x3=(-1)^(m+1)*arcsin3/4+πm,m-Z