Solve: #sinx+cosx=sqrt(2)# one the interval # (-2pi<=x<=2pi)# ?

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Solve: #sinx+cosx=sqrt(2) (-2pi<=x<=2pi)#

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Answer 1 · 2018-04-09T17:52:01

#x=pi/4, (3pi)/4, -pi/4, -(3pi)/4#

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#sinx+cosx=sqrt2#

#(sinx+cosx)^2=2#

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

#1+2sinxcosx=2#

#2sinxcosx=1#

We have a double angle identity that give us:

#sin2x=2sinxcosx#

#sin2x=1#

#2x=arcsin(1)=pi/2, (3pi)/2, -pi/2, -(3pi)/2#

#x=pi/4, (3pi)/4, -pi/4, -(3pi)/4#