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

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

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Answer 1 · 2018-03-11T05:57:35

$x_{1} = 0$ $x_1=0$ , $x_{2} = \frac{π}{3}$ $x_2=pi/3$ , $x_{3} = π$ $x_3=pi$ , $x_{4} = \frac{5 π}{3}$ $x_4=(5pi)/3$ 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 $f (x) = sin 2 x + sin x$ $f(x)=sin 2x+sin x$ (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,
$sin 2 x = 2 sin x \cdot cos x$ $sin 2x=2sin x* cos x$

Therefore
$2 sin x \cdot cos x - sin x = 0$ $2sin x* cos x- sin x=0$

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

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

Referring to a unit circle, along with $arcsin$ $arcsin$ and $arccos$ $arccos$ functions on your calculator if necessary, and we find
$x_{1} = 0$ $x_1=0$ , $x_{2} = π$ $x_2=pi$ , and
$x_{3} = \frac{π}{3}$ $x_3=pi/3$ , $x_{4} = \frac{5 π}{3}$ $x_4=(5pi)/3$ .

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

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

Answer 2 · 2018-03-11T05:59:09

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

Explanation:

Sin double angle identity:
$sin 2 x = 2 sin x cos x$ $Sin2x=2SinxCosx$

$2 sin x cos x - sin x = 0$ $2SinxCosx-sinx=0$
Factor with GCF:
$sin x (2 cos x - 1) = 0$ $sinx(2cosx-1)=0$
$sin x = 0$ $sinx=0$
$2 cos x - 1 = 0$ $2cosx-1=0$
$cos x = \frac{1}{2}$ $cosx=1/2$

You won't need inverse trig as these values are on the unit circle-
For $sin x = 0$ $sinx=0$
$x = 0, π (3.14)$ $x=0, pi (3.14)$
For $cos x = \frac{1}{2}$ $cosx=1/2$
$x = \frac{π}{3} (1.05), \frac{5 π}{3} (5.24)$ $x=pi/3 (1.05), (5pi)/3(5.24)$

Answer 3 · 2018-03-11T06:01:38

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
$(sin (2 x) - sin (x) = 0, x) ∣ 0 \leq x < 2 π$ $(sin(2x)-sin(x)=0,x)| 0<=x<2pi$

$x \in {0, \frac{π}{3}, π, 5 \cdot \frac{π}{3}}$ $x in {0, pi/3, pi, 5*pi/3}$

Into decimals:
0, 1.05, 3.14, 5.24

Answer 4 · 2018-03-11T20:49:13

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, $x = π$ $x = pi$ , and $x = 2 π$ $x = 2pi$ (rejected as outside of interval)
b. 2cos x - 1 = 0
$cos x = \frac{1}{2}$ $cos x = 1/2$
Trig table and unit circle give 2 solutions;
$x = \frac{π}{3}$ $x = pi/3$ , and $x = \frac{5 π}{3}$ $x = (5pi)/3$
Answers for half closed interval [0, 2pi):
$0, \frac{π}{3}; π; \frac{5 π}{3}$ $0, pi/3; pi; (5pi)/3$
In radian:
[0, 1.05, 3.14, 5.24) -> Answer # 4

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

4 Answers

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