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Abdallah2012
tan theta + sec theta = 1 between 0 and 2 pi
can theta=0 ? Like is is\[ 0< \theta <\Pi\] Or \[ 0 \leq \theta \leq \Pi\] ?
0<θ<2Π has to be between this interval
use substitution y=tan(theta/2) |dw:1383969602615:dw|
0≤θ<2Π sorry correction, it has to be between this interval
with a little bit of work you will find the solution
@Mertsj what about if they have a log like for example using the graph solve the system y=-2x+3 and y=logx+1 im confused cause of the log
@RED123url I am not sure what you did there...
well y=tan(theta/2) sin(thata)= 2y/(y^2+1) cos(theta)=(1-y^2)/(y^2+1) you know tan=sin/cos sec=1/cos plug things in and you will get an equation of y then go back y=tan)theta/2) and use the inverse
The answer is 0 but I used trial and error
@RED123url so just to confirm what swissgirl got, did you get the same answer?
I graphed it to confirm my answer and its def right
Well it is 0 and 2 pi But if you noticed the interval is \[ 0 \leq \theta < 2\pi \]
@RED123url well 2pi is not one the answers but 0 is, also according to the interval 2pi cannot be the answer..
the real answer for these kind of problem involves n: the answer is 2npi. n can be 0 or any number even a billion
@RED123url so would you agree that in this case the answer is 0?
if 2pi is excluded of course
well then. I am going with 0. Thank you both for your help!