How do you solve #cosx + 1 = sinx#?

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Answer 1 · 2015-04-16T14:57:42

This is a linear equation is sinus and cosine.

The faster way, to solve it, is this:

#sinx-cosx=1# let's multiply all the members with #sqrt2/2#

#sinx*sqrt2/2-cosx*sqrt2/2=sqrt2/2# and the first member is the development of: #sin(x-pi/4)#, so:

#sin(x-pi/4)=sqrt2/2rArr#

#x-pi/4=pi/4+2kpirArrx=pi/2+2kpi#

and

#x-pi/4=3/4pi+2kpirArrx=pi+2kpi#.