Solve this: #2sin2x+2sinx=2cosx+1#?

Question

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#

10623 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-06T15:09:22

See below.

So the part you missed was when you crossed out the #2cosx+1#. We must set that equal to zero as well--we cannot simply ignore it.

#2cosx+1=0#
#cosx=-1/2#
And we reach the solution you missed.

Answer 2 · 2017-04-06T15:11:26

Please see the explanation.

Given: #2sin(2x)+2sin(x)=2cos(x)+1#

You did this step:

#4sin(x)cos(x) + 2sin(x)= 2cos(x)+1#

At this point you should have subtracted #2cos(x)+1# from both sides:

#4sin(x)cos(x) + 2sin(x) - (2cos(x)+1) = 0#

Factor by grouping:

#2sin(x)(2cos(x)+1) - (2cos(x)+1) = 0#

#(2sin(x) -1)(2cos(x)+1) = 0#

#sin(x) = 1/2 and cos(x) = -1/2#

This will give your missing roots.