Trigonometric Equations

I found the solution to 61 online. I have a few questions about certain solution steps.



Here is the solution:



1. For part (a), Nghi N. said the unit circle gives 3 solutions for sin (4x) = 0.

4x = 0

4x = pi

4x = 2pi

Where on the unit circle is this found? I think the problem here is the fact I may not know how to properly read the famous unit circle. I don't understand why Nghi N. equated 4x to those 3 specific radian values.


2. The same for his work in part (b). I know how to solve cos(2x) = 0 for 2x (by taking the arccos on both sides of the equation). Nghi N. went on to say the unit circle gives 2 solutions:

2x = pi/2

2x = 3pi/2

I don't understand why he equated 2x to those two specific radian values. He ended up with 5 solutions on [0, 2pi). Please, explain.

 

you are looking for sin=0 and cos=0 on unit circle

4x = 0

4x = pi

4x = 2pi
so, first time sin=0 is at 0 (0° ), second time you go half circle to pi (180° ) , continue to make full circle which is 2pi (3600° )and you actually come to 0 (0° ) again

as you can see on unit circle, cos=0 only if angle is 90° or pi/2 and 270° or 3pi/2
then
2x = pi/2
2x = 3pi/2

 

Beautiful answer. Easy to understand.

 

Question 63

 

I don't feel good about this one.

 

yes

sin(x)=1/2

x = π/6
x = (5 π)/6

 

solutions:
x=0
x = π/4 and x = (3 π)/4
x = π/2
x = π

 

I was right with the exception that I forgot to include 3pi/4.