Solve: e^(2sinx)=1 , for 0≤x≤16
I'm not sure how to approach this problem. Can anyone explain how I should solve this?
Apply Ln to both sides - i.e.
Ln[e^(2sinx)] = Ln[1] , for 0≤x≤16
Please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.
.Well, I understand that you have to solve it but what does the limit mean exactly? (0≤x≤16) D0es this mean I have to plug it into the remaining equation:
2sin(x) = 0 → sin(x) = 0 ...... Now consider for what values of 'x' - you get sin(x) = 0
like this?: 2sin0= or 2sin16=
I'm not sure how i'll get a valid answer...Or do i have to use the unit circle?