b)
3-3cosx=2(< var > 2cos^{2}(x) < /var ><var>2cos
2
(x)</var> -1)
4< var > cos^{2}(x)+3*cos(x)-5=0 < /var ><var>cos
2
(x)+3∗cos(x)−5=0</var> cos(x)=t,-1<t<1
4t^2+3t-5=0
D=9+80=89
< var > \sqrt{D}=89\approx=9.4 < /var ><var>
D
=89≈=9.4</var>
< var > x_1=(-3-9.4)/8=-5/8 < /var ><var>x
1
=(−3−9.4)/8=−5/8</var>
< var > x_2=(-3+9,4)/8=0.8 < /var ><var>x
2
=(−3+9,4)/8=0.8</var>
Обратная замена
cosx=-5/8 cosx=0.8
x=+-arccos(-5/8)+2< var > \pi < /var ><var>π</var> *n x=+-arccos(0.8)+2< var > \pi < /var ><var>π</var> *n
