Значит, область значений на этом промежутке:
{z+3>3z-1⇒3z-z<3+1⇒2z<4⇒z<2
{5z-1>6-2z⇒5z+2z>6+1⇒7z>7⇒z>1
{z-3<0⇒z<3
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x∈(1;2)
{0,7(1+5z)<0,5(z+1)+3z⇒0,7+3,5z<0,5z+0,5+3z⇒3,5z-3,5z<-0,7 нет решен
{2z-(2-1,7)>6,7⇒2z-0,3>6,7⇒2z>7⇒z>3,5
x∈∅
если надо упростить .то начнем с 1+cos2x
1+cos2x = sin²x + cos²x + cos²x - sin²x = 2cos²x
tgx = sinx/cosx
sinx/cosx * 2*cos²x = 2 * sinx * cosx = sin2x
Решение
(а-6)² - (а+5)² = (a-6+a+5)(a-6-a-5) = -11(2a-1)