Trigonometric Equations

[nycmathguy]

Section 5.4

Can you help me with 69 and 71?

 

[MathLover1]

69.

sin(x+pi)-sin(x)+1=0........... [0,2π)

sin(x+pi)=-sin(x)

-sin(x)-sin(x)+1=0

1-2sin(x)=0
1=2sin(x)
sin(x) =1/2
x =sin^-1(1/2)
x=π/6
general solutions:

x=π/6+2π*n

solutions in interval [0,2π):

Radians:
x=π/6
x=5πi/6

Degrees:
x=30 °
x=150 °


71.

cos(x+π/4)-cos(x-π/4)=1 .......... [0,2π)

=>cos(x+π/4)=cos(x)/sqrt(2) - sin(x)/sqrt(2)
=> cos(x-π/4)=sin(x)/sqrt(2) + cos(x)/sqrt(2)

then
cos(x)/sqrt(2) - sin(x)/sqrt(2) -(sin(x)/sqrt(2) + cos(x)/sqrt(2))=1
cos(x)/sqrt(2) - sin(x)/sqrt(2) -sin(x)/sqrt(2) - cos(x)/sqrt(2))=1.......simplify
-sqrt(2) sin(x)=1
sin(x)=1/-sqrt(2)
x=sin^-1(1/-sqrt(2))
x=-π/4

general solutions:
x=-π/4+2pi*n
Solutions for\ the range : 0<=x<2pi

n=1
x=-π/4+2pi
x = (7π)/4->315° (degrees)

no solution for x in given interval [0,2π)

 

[nycmathguy]

69.

sin(x+pi)-sin(x)+1=0........... [0,2π)

sin(x+pi)=-sin(x)

-sin(x)-sin(x)+1=0

1-2sin(x)=0
1=2sin(x)
sin(x) =1/2
x =sin^-1(1/2)
x=π/6
general solutions:

x=π/6+2π*n

solutions in interval [0,2π):

Radians:
x=π/6
x=5πi/6

Degrees:
x=30 °
x=150 °


71.

cos(x+π/4)-cos(x-π/4)=1 .......... [0,2π)

=>cos(x+π/4)=cos(x)/sqrt(2) - sin(x)/sqrt(2)
=> cos(x-π/4)=sin(x)/sqrt(2) + cos(x)/sqrt(2)

then
cos(x)/sqrt(2) - sin(x)/sqrt(2) -(sin(x)/sqrt(2) + cos(x)/sqrt(2))=1
cos(x)/sqrt(2) - sin(x)/sqrt(2) -sin(x)/sqrt(2) - cos(x)/sqrt(2))=1.......simplify
-sqrt(2) sin(x)=1
sin(x)=1/-sqrt(2)
x=sin^-1(1/-sqrt(2))
x=-π/4

general solutions:
x=-π/4+2pi*n
Solutions for\ the range : 0<=x<2pi

n=1
x=-π/4+2pi
x = (7π)/4->315° (degrees)

no solution for x in given interval [0,2π)

Thank you. I will do 2 or 3 additional trig equations when time allows. As you know, my weekend is over.