Average value of a function on an interval. Will post equstion and work in a comment

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Average value of a function on an interval. Will post equstion and work in a comment

Mathematics
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\[\frac{ 1 }{ \pi - 0 } \int\limits\limits_{0}^{\pi} \sin (x) dx\] \[\int\limits_{0}^{\pi} \sin(x) dx = [-\cos] _{0}^{\pi}\] \[[-\cos \pi] - [-\cos 0] => [-(-1)] - [-1] = 1 + 1 = 2\] Average value = \[\frac{ 1 }{ \pi } (2) = \frac{ 2 }{ \pi }\] I'm asked to find the x value that corresponds to 2/pi sin(x) = 2/pi using the arcsin on the calculator I get ~ 0.690 But there is a second x value that corresponds to 2/pi, which is 2.451, how do I solve for the 2nd x value?
Can I find the 2nd x value algebraically, using calculus, or using a graphing calculator?

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Other answers:

it's going to be symmetrical about \(\pi/2\)|dw:1443181191167:dw|
@IrishBoy123 Thank you. How do I calculate it?
|dw:1443192745454:dw|
|dw:1443192978309:dw|
Ok. This is what I did. Please let me know if I over complicated it, or if there's a simpler way. Since the interval is [0, pi] and the first co-ordinate is 0 + arcsin(2/pi) the second co-ordinate is pi - arcsin(2/pi) = 2.451
seems fair enough
@IrishBoy123 How did you calculate it?
the calculator gives you the first answer, but you know from the curve that there are two. the symmetry of the sine curve gives you the second, which was what i was getting at with this: |dw:1443210395870:dw| but from knowing \(\sin(\pi - x) = \sin \pi \cos x - \sin x \cos \pi = \sin x\) you migt hav egot a result from another direction but i saw the symmetry of the curve....if that answers your question
I get the symmetry of the sine curve. I don't understand sin(π−x)=sinπcosx−sinxcosπ=sinx
second part was just saying/showing that \(sin (\pi - x) = \sin x\)
ok thanks for your help

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