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.

## Answer this Question

## Related Questions

- Precalculus(NEED HELP ASAP PLEASE!!) - Solve: sin(theta)cos(theta)-sin(theta)&#...
- Precalculus(NEED HELP ASAP PLEASE!!) - cot(theta)= 3 pi < theta < 3pi/2 ...
- Precalculus(NEED HELP ASAP PLEASE!!) - PLEASE HELP!! csc(theta)= -3/2 3pi/2 <...
- Trigonometry - Thank you STEVE!!!!! Suppose that Cos (theta) = 1/square root 2 ...
- Calculus - I wanted to confirm that I solved these problems correctly (we had to...
- Precalculus 2 - Similiarly to a question I asked previously I took the same ...
- Precalculus(NEED HELP ASAP PLEASE!!) - PLEASE HELP!! sin(theta)= 1/3 pi/2 < ...
- Precalculus check answers help! - 1.) Find an expression equivalent to sec theta...
- Precalculus check answers help! - 1.) Find an expression equivalent to sec theta...
- Precalculs - I have no idea how to do these type of problems. -------Problem...