Posted by **kristie** on Tuesday, August 1, 2006 at 3:51pm.

I have to find an exact solutions on the interval [0,pi) for both of these problems--

1) 2sin2x = sqrtx and

2) 1-2sqrt2 sinxcosx = 0

I know this much:

1) sin2x = (sqrt x)/2

= 2sinxcosx = (sqrt x)/2

What do I need to do next?

2) 1/(2 sqrt2) = sinxcosx

= (sqrt2)/4 = sinxcosx

= 2(sqrt2)/4 = 2sincosx

= (sqrt2)/2 = sin2x

And what do I do next?

Thanks!

2) 1/(2 sqrt2) = sinxcosx

= (sqrt2)/4 = sinxcosx

= 2(sqrt2)/4 = 2sincosx

= (sqrt2)/2 = sin2x

And what do I do next?

sin2x= .707

2X= 45 principle angle, 135 second

solve for x

I didn't check your math above.

1) sin2x=1/2 * sqrt(x) my mind is running a blank on this one. If I can see the solution, I will repost.

I still don't understand- What did you mean by 135 second?

the sin of 45 is the same as 90+45...it is symettrical about the 90 degree point.

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