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How do you solve #5sinx+2=sinx# for #0<=x<=2pi#?

Answer 1

#x in {pi/6,(11pi)/6}#

We start as if this wasn't a trigonometric equation: #5sinx+2=sinx# For now we'll 'pretend' that its domain is #D_0=\mathbb{R}# (all #x#s that make sense) and at the end we'll adjust the solutions to its actual domain #D=[0,2pi]#.

#5sinx+2=sinx# #5sinx-sinx=-2# #4sinx=-2# #sinx=-2/4# #sinx=-1/2#

Now, here we have at least two ways to go, I personally would use the fact that sine is an odd function:

#-sinx=1/2# #sin(-x)=1/2#

#-x=pi/6+2kpi or -x=pi-pi/6+2kpi, k in mathbbZ# #x=-pi/6+2kpi or x=-(5pi)/6+2kpi, k in mathbbZ# (Note: during the last step we multiplied both equations by #(-1)# and yet #+2kpi# stayed #+2kpi# - didn't change to #-2kpi#. The reason for that it that here it doesn't really matter because #k# goes trough the set of all integers - both positive and negative. I'm just used to writing #+somepi#)

Now we can adjust our solution to the domain #D=[0,2pi]#. Concluding: #x in {(7pi)/6,(11pi)/6}#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you solve #5sinx+2=sinx# for #0&lt;=x&lt;=2pi#?

How do you solve #5sinx+2=sinx# for #0<=x<=2pi#?