Multiple-Angle Equation...2

Section 5.5







Correction

Answer:

pi/2 + 2pi•k

-pi/4 + pi•k

 

x=(π*k)/2, k element Z
x = (π k)/2 - π/4, k element Z

 

Explain how you came up with the answers as posted.

 

(sin(2x)+cos(2x))^2=1
(sin(2x)+cos(2x))^2-1=0
(sin(2x)+cos(2x)+1)(sin(2x)+cos(2x)-1)=0

(2sqrt(2) sin(x + π/4) cos(x))(2 sqrt(2) sin(π/4 - x) sin(x))=0

make each factor =0
2sqrt(2) sin(x + π/4) cos(x)=0 if sin(x + π/4) =0 or cos(x)=0
and
2 sqrt(2) sin(π/4 - x) sin(x))=0 if sin(π/4 - x)=0 or sin(x)=0

sin(x + π/4) =0 for x = π n +π/4, n element Z
cos(x)=0 for x = π n - π/2, n element Z
sin(π/4 - x)=0 for x = π n + (3π)/4, n element Z
sin(x)=0 for x = π n, n element Z -> Integer solution is x=0

general solutions are:
x= π*n
x=π*n+π/4
x=π/2+2π*n
x=3π/4+π*n

 

I was totally wrong. Not good, not good at all. Tell me, what's wrong with taking the square root on both sides as step 1?

 

sin(2x)+cos(2x)=sqrt(1)

you did this correct
2sin(x)cos(x)-2sin^2(x)=0
2(sin(x)cos(x)-sin^2(x))=0..........to make it simple, here you need to factor out sin(x)
2sin(x)(cos(x)-sin(x))=0

if 2sin(x)=0 =>x = π n, n element Z
or
if cos(x)-sin(x)=0->cos(x)=sin(x) =>x = (1/4 )(4π n + π), n element Z or π n+π/4

 

I will surely do a few more before moving on in the section.