Solving Trigonometric Equations...1

[nycmathguy]

Section 5.3

For reasons unclear to me, I have trouble solving trigonometric equations. I tried reading the lecture notes by the author but it is just not sinking into my brain. In that case, I would like someone, anyone, to do 20 and 24. If it's easier to use the unit circle to find values of x, please say so. Thanks.

 

[MathLover1]

20.

(2sin^2(x)-1)(tan^2(x)-3)=0.........will be zero if

(2sin^2(x) - 1) (tan^2(x)-3)=0

solutions:

if 2sin^2(x) - 1 =0 => 2sin^2(x)= 1 => sin^2(x)= 1/2 => sin(x)= sqrt(1/2 )

x=pi/4+2pi* n, x=3pi/4+2pi* n,x=5pi/4+2pi* n,x=7pi/4+2pi* n

if tan^2(x)-3=0 =>tan^2(x) = 3=>tan(x) = sqrt(3)

x=pi/3+pi*n, x= 2pi/3+pi*n

combine solutions:
pi/4+2pi* n
3pi/4+2pi* n
5pi/4+2pi* n
7pi/4+2pi* n
pi/3+pi*n
2pi/3+pi*n


24.
sec(x)csc(x)=2csc(x)
sec(x)csc(x)-2csc(x) =0
(sec(x)-2)csc(x) =0

solutions

if csc(x) =0 => there is no solution

if sec(x)-2 =0 =>sec(x)=2
x=pi/3+2pi*n,
x=5pi/3+2pi* n

 

[nycmathguy]

20.

(2sin^2(x)-1)(tan^2(x)-3)=0.........will be zero if

(2sin^2(x) - 1) (tan^2(x)-3)=0

solutions:

if 2sin^2(x) - 1 =0 => 2sin^2(x)= 1 => sin^2(x)= 1/2 => sin(x)= sqrt(1/2 )

x=pi/4+2pi* n, x=3pi/4+2pi* n,x=5pi/4+2pi* n,x=7pi/4+2pi* n

if tan^2(x)-3=0 =>tan^2(x) = 3=>tan(x) = sqrt(3)

x=pi/3+pi*n, x= 2pi/3+pi*n

combine solutions:
pi/4+2pi* n
3pi/4+2pi* n
5pi/4+2pi* n
7pi/4+2pi* n
pi/3+pi*n
2pi/3+pi*n


24.
sec(x)csc(x)=2csc(x)
sec(x)csc(x)-2csc(x) =0
(sec(x)-2)csc(x) =0

solutions

if csc(x) =0 => there is no solution

if sec(x)-2 =0 =>sec(x)=2
x=pi/3+2pi*n,
x=5pi/3+2pi* n

1. What is 2•pi•n?

2. Why so many solutions?

3. How can I use the unit circle to find all my solutions?

 

[MathLover1]

2pi is period, 2pi•n where n = 0, ±1, ±2, …. , and that way you find all possible solutions
So, if I wanted to solve sin x = 0, I would take the inverse sine of 0. I get an answer of 0 on my calculator, but in reality, sin x = 0 has an infinite number of solutions. As you can see on the graph, I get 0 for sin x every 2pi spaces. So my answers are actually x = 0 + 2*pi*n where n stands for the number of extra solutions that we have. N increases by 1 each time. We get x = 0 + 2*pi*0 = 0 for our first solution. The next solution is x = 0 + 2*pi*1 = 2pi. The next one is x = 0 + 2*pi*2 = 4pi. And so on and so forth.
The sine and cosine functions both have a period of 2pi. This means that your answers are 2pi spaces apart. The tangent function, on the other hand, has a period of pi. This means that these answers are pi spaces apart.

You can use unit circle too, but you also need to use a period.

 

[nycmathguy]

2pi is period, 2pi•n where n = 0, ±1, ±2, …. , and that way you find all possible solutions
So, if I wanted to solve sin x = 0, I would take the inverse sine of 0. I get an answer of 0 on my calculator, but in reality, sin x = 0 has an infinite number of solutions. As you can see on the graph, I get 0 for sin x every 2pi spaces. So my answers are actually x = 0 + 2*pi*n where n stands for the number of extra solutions that we have. N increases by 1 each time. We get x = 0 + 2*pi*0 = 0 for our first solution. The next solution is x = 0 + 2*pi*1 = 2pi. The next one is x = 0 + 2*pi*2 = 4pi. And so on and so forth.
The sine and cosine functions both have a period of 2pi. This means that your answers are 2pi spaces apart. The tangent function, on the other hand, has a period of pi. This means that these answers are pi spaces apart.

You can use unit circle too, but you also need to use a period.

Thanks. We'll see how We'll I do in this section.