Question

find the real solutions for the following quadratic equation: x^2+6x+3=0

Answer 1

\[\large \sin^{-1}(\sin \frac{7 \pi}{5}) \implies \color{green}{ \frac{7 \pi}{5}}\]

Answer 2

did they tell you a domain restriction?

Answer 3

What is the exact value of the following expression: sin^-1 (sin 11pie/8)

Answer 4

Looking at the last one, what do you think the answer is?

Answer 5

this is a function that is the inside of itself, what happens? f(f^(-1)(x)) = x

Answer 6

well the anwer is 7pie/5 but it had a range restriction of -pie/2, pie/2

Answer 7

does the second one?

Answer 8

yes it also has the same restrictions

Answer 9

the range restriciton makes it "go both ways" so that the inverse is also a fuction

Answer 10

then its the same deal

Answer 11

but according to the exact value it's not right
because it can't be bigger than 1

Answer 12

o wait I read that wrong 1 sec

Answer 13

ok

Answer 14

ok the first one is wrong also, sorry.
sin(7pi/5) is in the 3rd quad, but that is not within the domain of sin so when we go back it gives you the angle that is in the 4th quad

Answer 15

so sin^(-1)(sin(7pi/5)) = the anlge that is the same y value as 7pi/5 but in the 4th quad

Answer 16

make sense?

Answer 17

so take 7pi/5 and minus pi from it. this is how much you need to subtract from 0 to get the angle
so the answer is -(7pi/5-pi) = sin^(-1)(sin(7pi/5)) = -2pi/5 = aprx -1.25664

Answer 18

and 11pi/8 is the same deal
sin^(-1)(sin(11pi/8)) = -(11pi/8-pi)

Answer 19

wow