9х-6х+6=5х+10
9х-6х-5х=10-6
-2х=4
х=-2
Решение:
3tg^2x/3-2tgx/3-1=0
tgx/3=t
3t^2-2t-1=0
t=1 tgx/3=1 x/3=П/4+Пk x=3П/4+3пk
t=-1/3
tgx/3=-1/3 x/3=-arctg1/3+Пk
x=-3arctg1/3+3Пk.
<span><span><span>Применим формулу замены произведения косинусов их суммой
cos α · cos β</span>
=
(cos(α-β) + cos(α+β))/
</span><span>
2</span></span>
cos44*cos16=1/2(cos60+cos28)=1/2(1/2+cos28), cos60=1/2
cos 59*cos31=1/2(cos90+cos28)=1/2<span>cos28, </span> cos90=0
1/2(1/2+cos28)-1/2<span>cos28=1/4+1/2</span>cos28-1/2cos28=1/4