Trig helppp
Part I: Would my answers be theta = 3pi/4 and theta = 7pi/4 ?
`Part I:` `Would my answers be theta = 3pi/4 and theta = 7pi/4 ?` yes that's correct
Awesome! For Part II, I got: 3pi/4 + 2npi and 7pi/4 + 2npi Yeah?
while that is correct, there is a simpler way to condense those two general solutions you can write them both as x = 3pi/4 + pi*n where n is any integer
notice if n = 1, then x = 3pi/4 + pi*n x = 3pi/4 + pi*1 x = 3pi/4 + pi x = 3pi/4 + 4pi/4 x = 7pi/4
Oh that makes sense... how would I recognize to condense it?
It's just something you practice over and over til you get good at recognizing the different forms. I guess you could ask yourself: is one solution based off another? If so, then how? In this case, 7pi/4 is one half-turn from 3pi/4, which is why you add on pi. To get the next solution, you add on another pi, and so on. Which is why the `pi*n` portion is there.
But 3pi/4 + 2npi and 7pi/4 + 2npi works fine just as well
Awesome, thanks! That makes a lot of sense :)
I'm glad I could help
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