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 $x=4/3pi+2kpi$

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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.