Question

Not sure how they got those x's

Answer 1

you're substituting the x's in.

Answer 2

oh, in the original equation ?

Answer 3

so, 2sinx+sin^2x ...2sin(o)+sin^2(0) ? :S

Answer 4

Sorry could you please show me Im not very good at trig

Answer 5

You just posted it! Yes lol...

Answer 6

Okayyy so then 0+sin^2 (0) = 0 ?

Answer 7

2sin(-1) + sin^2 (-1) = I got decimals

Answer 8

You got decimals? Then you're fine.

Answer 9

but I dont know how they got x= pi / 2 + 2n pi and 3pi/2 + 2npi

Answer 10

how is ling defined?

Answer 11

oh tangent line=ling?

Answer 12

\[\cos(x)=0 \text{ when } x=\pi/2 \text{ or } x=3 \pi/2 \text{ right? } \]

Answer 13

This is according to the unit circle

Answer 14

now +2npi just means the number of rotations we went around a circle

Answer 15

ohhhhhhhhhh okay okay ! so then to find by y value. I would pi/2 and 3pi/ 2? so 2sin(pi/2)+sin^(pi/2) and 2sin(3pi/2)+sin^2(3pi/2) ?

Answer 16

\[f(\frac{\pi}{2}+2 n \pi )=2 \sin(\frac{\pi}{2}+2 n \pi )+[\sin(\frac{\pi}{2}+2 n \pi )]^2 \\ = 2[\sin(\frac{\pi}{2})\cos(2\pi n)+\cos(\frac{\pi}{2})\sin(2 \pi n)] \\+[(\sin(\frac{\pi}{2})\cos(2\pi n ) +\cos(\frac{\pi}{2})\sin(2 \pi n)]^2 \\ =2[1 \cdot \cos(2 \pi n) + 0 \cdot 0 ]+[1 \cdot \cos(2 \pi n)+0 \cdot 0 ]^2 \]
\[=2 \cos(2 \pi n )+(\cos(2 \pi n ))^2 \]

so cos(2pi)=1 so cos(2pi*n)=1

Answer 17

and yeah you could have just plugged in pi/2
but i wanted to by fancy

Answer 18

and also plug in 3pi/2 also