Trigonometric Equations...3

[nycmathguy]

Section 5.4

Question 74

sin(x + pi/2) - cos^2 x = 0

Let sin(x + pi/2) = cos x

cos x - cos^2 x = 0

Factor

cos x(1 - cos x ) = 0

Set each factor to 0 and solve for x.

cos x = 0

x = pi/2, 3pi/2

Set the other factor to 0 and solve for x.

1 - cos x = 0

-cos x = -1

cos x = -1/-1

cos x = 1

x = 2pi

You say?

 

[Country Boy]

I would not say "Let sin(x + pi/2) = cos x". That is an identity, you do not need to "let" it be true in this particular case.

 

[nycmathguy]

I would not say "Let sin(x + pi/2) = cos x". That is an identity, you do not need to "let" it be true in this particular case.

Is my answer correct?

 

[Country Boy]

cos(0)= 1 so x= 0 is an solution. If you want all solutions, since sine and cosine are periodic with period 2pi, a multiples of 2p:
x= pi/2+ 2npi
x= 3pi/2+ 2npi
x= 2npi
for any integer, n.

 

[nycmathguy]

cos(0)= 1 so x= 0 is an solution. If you want all solutions, since sine and cosine are periodic with period 2pi, a multiples of 2p:
x= pi/2+ 2npi
x= 3pi/2+ 2npi
x= 2npi
for any integer, n.

I was right, then.