can i do a square root here?
assuming I cant I went on and did
[tex](4cos^4x-2cos^2x+1)+cos^2x=1/4[\math]
how .. do I do this.. especially without the unit circle
I looked at the answers and the values are not on the unit circle.
can i do a square root here?
assuming I cant I went on and did
[tex](4cos^4x-2cos^2x+1)+cos^2x=1/4[\math]
how .. do I do this.. especially without the unit circle
I looked at the answers and the values are not on the unit circle.
does NOT equal .
So far you have two possibilities:
Case 1:
.
Cosine is positive in the first and fourth quadrants, and to get to , you have to "undo" the cosine - so take the arccos (or ...
, since the period of cosine is .
.
Case 2:
, since cosine is negative in the second and third quadrants and the period is
.
So putting it all together, the possibilities are
.