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

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Answer 1 · 2016-12-28T16:20:45

The answer is # S={2kpi,pi/2+2kpi}#, #k in ZZ#

We need

#sin(A+B)=sinAcosB+sinBcosA#

#sin^2A+cos^2A=1#

We compare this equation to

#rsin(x+a)=1#

#rsinxcosa+rcosxsina=1#

#sinx+cosx=1#

Therefore,

#rcosa=1# and #rsina=1#

So,

#cos^2a+sin^2a=1/r^2+1/r^2=2/r^2=1#

#r^2=2#, #=>, #r=sqrt2#

and #tana=1#, #=>#, #a=pi/4#

Therefore,

#sqrt2sin(x+pi/4)=1#

#sin(x+pi/4)=1/sqrt2#

#x+pi/4=pi/4+2kpi#, #=>#, #x=2kpi#

and

#x+pi/4=3pi/4+2kpi#, #=>#, #x=pi/2+2kpi#

The solutions are # S={2kpi,pi/2+2kpi}#, #k in ZZ#

Answer 2 · 2016-12-28T16:33:47

Here's an alternative answer. Square both sides.

#(sinx + cosx)^2 = 1^2#

#sin^2x + 2sinxcosx + cos^2x = 1#

Use the identity #sin^2theta + cos^2theta = 1#.

#1 + 2sinxcosx = 1#

#2sinxcosx = 0#

Use the identity #2sinthetacostheta = sin2theta#:

#sin2x = 0#

#2x = 0, pi#

#x = 2pin, pi/2 + 2pin#, where #n# is an integer.

Hopefully this helps!