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

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4 Answers
Mar 11, 2018

#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.

Mar 11, 2018

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)#

Mar 11, 2018

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

Mar 11, 2018

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