Posted by **jh** on Thursday, March 25, 2010 at 7:35pm.

Solve:

sin(theta)+cos(theta) = 1

0 < or equal to theta < 2pi

Solution Set:__________ ?

- Precalculus(NEED HELP ASAP PLEASE!!) -
**Damon**, Thursday, March 25, 2010 at 8:34pm
sin T +sqrt (1-sin^2T) = 1

sqrt(1-sin^2T) = 1 - sin T

1 - sin^2 T = 1 - 2 sin t + sin^2 T

-sin^2 T = -2 sin T + sin^2 T

2 sin^2 T = 2 sin T

sin T = 1

T = 0

That tells you something. The only places this happens are when sin = 0 or cos = 0

That is at 0, pi/2, pi, and 3 pi/2

- Precalculus(NEED HELP ASAP PLEASE!!) -
**Damon**, Thursday, March 25, 2010 at 8:35pm
Oh, and then only when the one that is not zero is +1, not -1

- Precalculus(NEED HELP ASAP PLEASE!!) -
**jh**, Thursday, March 25, 2010 at 8:59pm
Damon I don't understand you 2nd post but is the answer: 0, pi/2, pi, and 3 pi/2 ??

If it is.. then unfortunately its not the correct answer. Do you think you went wrong somewhere?

- Precalculus(NEED HELP ASAP PLEASE!!) -
**Damon**, Thursday, March 25, 2010 at 9:06pm
sin T = +1 when T = 90 degrees , pi/2

cos T = 1 when T = 0 or 2pi

if T = 0 for example

sin T = 0

cos T = 1

and sin T + cos T = 1

So that just plain checks.

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