How do you solve #cos^2 theta-sin^2 theta +sin theta=0#?

1 Answer
Dean R.
Jun 4, 2018

#theta = -30^circ + 360^circ k quad# or

#theta = -150^circ + 360^circ k # or

#theta = 90^circ + 360^circ k #

all for integer #k#.

Explanation:

I think I just did this one recently.

#cos ^2 theta - sin ^2 theta + sin theta = 0#

#1 - sin ^2 theta - sin ^2 theta + sin theta = 0#

# 2 sin theta - sin theta -1 = 0#

#(2 sin theta +1 ) (sin theta - 1) = 0#

#sin theta = -1/2 or sin theta = 1 #

#theta = -30^circ + 360^circ k quad# or

#theta = -150^circ + 360^circ k # or

#theta = 90^circ + 360^circ k #

all for integer #k#.

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