Question #d1d01

1 Answer
Y
May 10, 2017

#x=pi/3, (5pi)/3, pi/2#

Explanation:

Formatted question: #(2cosx-1)(sinx-1)=0#

We can easily see that in this equation, #2cosx-1=0# or #sinx-1=0#. So, we can solve for x:
#2cosx-1=0#
#cosx=1/2#
#x=pi/3# or #x=(5pi)/3# using our special unit circle values

#sinx-1=0#
#sinx=1#
#x=pi/2# using our special unit circle values

Therefore, #x=pi/3, (5pi)/3, pi/2#

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