Question

5.Solve using the interval [0,2pi]
Part I: Using an appropriate inverse trigonometric expression, write an equation that defines the value of 3x. (2 points)
Part II: Solve this equation to find all possible values of the angle 3x. (4 points)
Part III: Use algebra to find all values of x between 0 and 2π that satisfy this equation. (4 points)

Answer 1

ou're looking for 3x.
You can solve this by using inverse cos.

So, 3x = arccos(1)
[***remember, arc = inverse = ^-1***]

So now that you know 3x = arccos(1),
use a calculator to solve for arccos(1).
arccos(1) = 0.

Answer 2

part 1

Answer 3

so 3x = 0?

Answer 4

0 is a reference angle. When is cosine equal to 0?
When the angle is π/2 and 3π/2 (you can solve this by trigonometry or looking at the when the cosine graph is 0),
So...
3x = π/2 ... and 3x = 3π/2

3x = π/2
x = π/6

3x = 3π/2
x = π/2

Answer 5

so how do i do parts 2& 3

Answer 6

part 2 is 43

Answer 7

part 3 .....
allow 3x to equal a

cos(a) = -1

a = 3(Pi)/2, 7(Pi)/2, 11(Pi)/2

3x = 3(Pi)/2, 7(Pi)/2, 11(Pi)/2

x = 3(Pi)/6, 7(Pi)/6, 11(Pi)/6

Answer 8

thanks

Answer 9

no problem
i here to help

Answer 10

i have more problems i need help with