Trigonometry Help

Question

Please describe your answer to the following question:

Q: Solve the given equation for x; find all solutions in the interval [0,2pi)

cos(x)+1=sin(x)

A step by step would be great to help me process the material better Please and Thank You!

Answer 1

cosx +1=sinx

sinx -cosx = 1

square both sides

sin^2(x) -2sinxcosx +cos^2(x)= 1

sin^2(x)+cos^2(x) -2sinxcosx=1

1-2sinxcosx=1

-2sinxcosx=0

sinxcosx =0

set each factor =0

sinx =0

cosx=0

x= 0, pi/2, pi, 3pi/2, 2pi

but some are extraneous solutions not in the original equation

which happens due to multiplying by a variable

check each one

throw out 0, 3pi/2 and 2pi

x= pi/2, pi

or, another strategy:

cosx +1=sinx= sqr(1-cos^2(x))

cosx +1 =sqr(1-cos^2(x))

square both sides

cos^2(x)+2cosx +1= 1-cos^2(x)

2cos^2(x)+2cosx =0

cosx(cosx+1)=0

set each factor =0

cosx=0 or -1

x =pi/2, 3pi/2,pi

throw out 3pi/2

x= pi/2, pi

Answer 2

here is useful tool for solving for angle

when the problem is:

Acos(x)-Bsin(x)=k

this can be rewritten as

√(A²+B²)[cos(x+tan^-1(B/A)]=k

derive saying A=rcos(ß),,,,B=rsin(ß)

obviously A²+B²=r²,,,,,B/A=tan(ß),,,,,,and rcos(ß)cos(x)-rsin(ß)sin(x)=rcos(x+ß)

this problem

cos(x)-sin(x)=-1

(√2)cos(x+tan^-1(1/1))=-1

cos(x+π/4)=-(√2)/2

x+π/4=(+/-)3π/4

x₁=(π/2)+2πk,,,k set to zero so x₁=π/2

x₂=-π+2πk,,,,k set to 1 so x₂=π

no extraneous solution points

note that f(x)=cos(x)-sinx

has same graph as g(x)=(√2)cos(x+π/4)