What are the x-intercepts of the graph of f(x) = 2 sin(3x) + rad3 on the interval [0, 2theta]?

Question

 I'm still new and confused with sin graphs. I know the answer is 4pi/9, 5pi/9, 10pi/9, 11pi/9, 16pi/9, 17pi/9 but unsure of how the denominator got to be 9.

 

thank you!

Answer 1

You probably forgot to notice the 3x in the sin function...

 

First of all, I assume you mean the interval [0, 2π], right?

 

An x-intercept means that y = 0.  So solve the equation:

 

2 sin(3x) + √3 = 0

to find the x-intercepts.

 

2 sin(3x) + √3 = 0

2 sin(3x) = -√3

sin 3x = -√3/2

 

For what values is the sin -√3/2?  That would be 4π/3 and 5π/3, and all values which are these numbers ± 2π.  So you can write 4π/3 ± 2πn and 5π/3 ± 2πn, where n is any integer.

 

So now you have the following:

 

sin 3x = -√3/2, so

3x = 4π/3 ± 2πn, or

3x = 5π/3 ± 2πn

 

Now you have to solve for x by dividing by 3

 

x = 4π/9 ± 2πn/3, or

x = 5π/9 ± 2πn/3

 

So x is 4π/9 or 5π/9, ± any multiple of 2π/3.

 

You can ignore the - part of the ± because as soon as you subtract 2π/3 from either 4π/9 or 5π/9, you get into negative values for x, which can't be right because the interval for x starts at 0.

 

So you just have to add multiples of 2π/3 to 4π/9 and 5π/9 until you get above 2π.