2sin²(-2490)+tan 1410=2sin²(-330-6*360)+tan (330+3*360)=
=2sin²(-330)+tan (330)=2 sin²(30-1*360)+tan (-30)=
=2sin²(30)-tan (30)=2(1/2)²-√3/3=1/2-√3/3)=(3-2√3)/6
Answer: 2 sin²(-2490)+tan (1410)=(3-2√3)/6 (≈-0.07735...)
Apply slip and slide
a^2-3a-4
(a-4)(a+1)
(a-2)(2a+1)
Find zeros
a-2=0
a=2
2a+1=0
2a=-1
a=-1/2
Final answer: {-1/2, 2}
Answer:21,849.24
Step-by-step explanation:
