Solve this: #2sin2x+2sinx=2cosx+1#?
Solve this: #2sin2x+2sinx=2cosx+1#
I tried, but something is bad.
#4sinxcosx+2sinx=2cosx+1#
#2sinx(2cosx+1)=2cosx+1#
#2sinx=1#
#sinx=1/2#
#x=pi/6+2kpi# or #x=(5pi)/6+2kpi#
I missed somewhere:
#x=2/3pi+2kpi# and #x=4/3pi+2kpi#
Solve this:
I tried, but something is bad.
I missed somewhere:
2 Answers
See below.
Explanation:
So the part you missed was when you crossed out the
And we reach the solution you missed.
Please see the explanation.
Explanation:
Given:
You did this step:
At this point you should have subtracted
Factor by grouping:
This will give your missing roots.