Если эти квадраты равны , тогда 2+a
1+tg²2a=1/cos²2a
tg2a=4/3
1+16/9=1/cos²2a
cos²2a=9/25
cos2a=3/5 cos2a=-3/5
cos²a=1+cos2a/2 = 4/5 cos²a=1+cos2a/2=1/5
sina==-√1-cos²a sina=+-√1-cos²a
sina=1/√5 sina=-1/√5 sina=2/√5 sina=-2/√5
tg2a=0
1+0=1/cos²2a
cos²2a=1
cos2a=1 cos2a=-1
cos²a=1+cos2a/2 = 1 cos²a=1+cos2a/2=0
sina==-√1-cos²a sina=+-√1-cos²a
sina=0 sina=1 sina=-1
<span>tg(2x+П/6) > -√3
</span>пk -п/3 < 2x+п/6 < пk +п/2 <span>k C Z
</span><span>пk -п/3 -п/6 < 2x< пk +п/2 - п/6
</span><span>пk -п/2 < 2x < пk +п/3
</span><span>пk/2 -п/4 < x < пk/2 +п/6, k C Z</span>
A) 2x3 - 7 + 4 + 3x2 - 5x3 = -3x3 + 3x2 - 3 = -3 (x3-x2+1)
б)z2 - 3z + 2 + 4z + 8 + 3z2 - 5 = 4z2 + z +5
в)3t3-4t2+7t +2t2 - 6t +7= 3t3-2t2+t+7
г)2a2 + 5a - a2 + a + a2 -3a-5 = 2a2+3a-5