Question #ab74b

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Answer 1 · 2017-07-31T17:45:53

#theta = pi/2, (2pi)/3, (4pi)/3#

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

#2sin theta cos theta - 2cos theta +sin theta -1=0 # #-># rearrange the terms

#2 cos theta (sin theta - 1) +1(sin theta -1)=0 # #->#factor

#(sin theta - 1)(2 cos theta +1) = 0# #-># factor out #sin theta - 1#

There are two possibilities:

#sin theta - 1 = 0#

#sin theta = 1#
#theta = pi/2#

#2 cos theta + 1 = 0#

#2 cos theta = -1#
#cos theta = -1/2#
#theta = (2pi)/3, (4pi)/3#

So, #theta = pi/2, (2pi)/3, (4pi)/3#.