Question

Use a calculator to solve the equation on the interval [0, 2π). (see picture for more info). Thanks a lot!?!

Answer 1

`x_1=0`, `x_2=pi/3`, `x_3=pi`, `x_4=(5pi)/3` and thus the last option.

Explanation:

Depending on the model you use, there can be a variety of approaches to find zeros on a particular interval. If you are using a GDC like the TI-84, you might be able to determine zeros of the equation by defining, plotting, and analyzing the graph of the function `f(x)=sin 2x+sin x` (which equals to the left-hand side of the equation).
http://www.dummies.com/education/graphing-calculators/how-to-find-the-zeroes-of-a-function-with-the-ti-84-plus/

On the other hand, you could have been able to solve this equation by applying the doubling angle identity for the sine function,
`sin 2x=2sin x* cos x`

Therefore
`2sin x* cos x- sin x=0`

Factor out `sin x`
`sin x(2cos x-1)=0`

By the factor theorem the function would have a zero as long as at least one of these equation holds:
`sin x=0`
`cos x=1/2`.

Referring to a unit circle, along with `arcsin` and `arccos` functions on your calculator if necessary, and we find
`x_1=0`, `x_2=pi`, and
`x_3=pi/3`, `x_4=(5pi)/3`.

Evaluate these expressions on your calculator and ask for the decimal output to find the answer choice to this question. (Use `pi=3.14` if you are calculating by hand.)

You can verify these results by substituting the equation with the respective values of `x`. Alternatively, you can trace the graph to see if you get an `x` -intercept at these points.

Answer 2

Alright what you plug into your calculator will be inverse trig...
See below

Explanation:

Sin double angle identity:
`Sin2x=2SinxCosx`

`2SinxCosx-sinx=0`
Factor with GCF:
`sinx(2cosx-1)=0`
`sinx=0`
`2cosx-1=0`
`cosx=1/2`

You won't need inverse trig as these values are on the unit circle-
For `sinx=0`
`x=0, pi (3.14)`
For `cosx=1/2`
`x=pi/3 (1.05), (5pi)/3(5.24)`

Answer 3

The answer is the last option
0, 1.05, 3.14, 5.24

Explanation:

Because the domain given lists 0 as inclusive, the 0 stays as a solution

I've plugged into my calculator

solve
`(sin(2x)-sin(x)=0,x)| 0<=x<2pi`

`x in {0, pi/3, pi, 5*pi/3}`

Into decimals:
0, 1.05, 3.14, 5.24

Answer 4

Answer #4

Explanation:

sin 2x - sin x = 0
Using trig identity: sin 2x = 2sin x.cos x, we get:
2sin x.cos x - sin x = 0
sin x.(2cos x - 1) = 0
Either factor should be zero.
a. sin x = 0
Unit circle gives -->
x = 0, `x = pi`, and `x = 2pi` (rejected as outside of interval)
b. 2cos x - 1 = 0
`cos x = 1/2`
Trig table and unit circle give 2 solutions;
`x = pi/3`, and `x = (5pi)/3`
Answers for half closed interval [0, 2pi):
`0, pi/3; pi; (5pi)/3`
In radian:
[0, 1.05, 3.14, 5.24) -> Answer # 4