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Solution of Trigonometric Equations

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  • #1
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Homework Statement


Find the solution set of each of the following equations for the interval 0° ≤ x ≤ 360°.
[tex]sin^2x = sinx[/tex]

Homework Equations


Trigonometric Identities for Sine.

The Attempt at a Solution


This is my attempt so far:

[tex]\sin^2x = sinx[/tex]
[tex]\sin^2x - sinx = 0[/tex]
[tex]\sin^2x - sinx + \frac{1}{4} = \frac{1}{4}[/tex]
[tex](sinx - \frac{1}{2})^2 = \frac{1}{4}[/tex]
taking square roots of both sides:
[tex]sinx -\frac{1}{2} = \frac{1}{2}[/tex]
[tex]sinx = ±\frac{1}{2} + \frac{1}{2}[/tex]
if [itex]\frac{1}{2},[/itex]
[tex]sinx = 1[/tex]
if [itex]-\frac{1}{2},[/itex]
[tex]sinx = 0[/tex]
take arcsin of both sides:
[tex]x = arcsin1[/tex]
[tex]x = 90° or \frac{\pi}{2}[/tex]
[tex]x = arcsin0[/tex]
[tex]x = 0°[/tex]

Solution set = {0°, 90°} or {0, [itex]\frac{\pi}{2}[/itex]}

EDIT: The complete solution set must be:

x = {0°, 90°, 180°, 360°}
because sinx = 0 also in 180 and 360, not only in 0°)
Special thanks to Sourabh N for guiding me to the correct solution o:)
 
Last edited:

Answers and Replies

  • #2
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Find the solution set of each of the following equations for the interval 0° ≤ x ≤ 360°.
[tex]sin^2x = sinx[/tex]


take arcsin of both sides:
[tex]x = arcsin1[/tex]
[tex]x = 90° or \frac{\pi}{2}[/tex] What other value(s) of x satisfies this equation?
[tex]x = arcsin0[/tex]
[tex]x = 0°[/tex] What other value(s) of x satisfies this equation?

Solution set = {0°, 90°} or {0, [itex]\frac{\pi}{2}[/itex]}
Notice the range of x: 0° ≤ x ≤ 360°. What other values of x within this range satisfy the conditions x = arcsin1 or x = arcsin0?
 
  • #3
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Notice the range of x: 0° ≤ x ≤ 360°. What other values of x within this range satisfy the conditions x = arcsin1 or x = arcsin0?
How about this?

x = {0, 90°, 360°}
 
  • #4
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No.

Here's a very good way to figure out all the solutions. Draw the curve sin[x] for 0° ≤ x ≤ 360°. Also draw horizontal lines at y = 0 and y = 1. The point(s) where sin[x] and these lines intersect, are your solutions.
 
  • #5
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No.

Here's a very good way to figure out all the solutions. Draw the curve sin[x] for 0° ≤ x ≤ 360°. Also draw horizontal lines at y = 0 and y = 1. The point(s) where sin[x] and these lines intersect, are your solutions.
Oh. You mean, what values of x does give the angles 0, 90°, 360° when worked with sine?
 
  • #6
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for x values,

x = {0, 1}
 
  • #7
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No. x IS your angle.

y = Sin[x].
x = x.

1. Draw y = 0. See at what values of x do Sin[x] and y = 0 intersect.
2. Same with y = 1.
 
  • #8
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No. x IS your angle.

y = Sin[x].
x = x.

1. Draw y = 0. See at what values of x do Sin[x] and y = 0 intersect.
2. Same with y = 1.
Is my graph correct?

2w3tmc7.png
 
  • #9
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On the horizontal axis, you have 0 -> 90 -> 360?? In 0->90-> What comes after 90?
 
  • #10
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This?
2w3wzsl.png
 
  • #11
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Very good! You have also (correctly) marked the points of intersection, except that you forgot to include x = 0.

So what is the solution set?
 
  • #12
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Very good! You have also (correctly) marked the points of intersection, except that you forgot to include x = 0.

So what is the solution set?
Yeah, and also the point at 360. Sorry :shy:

Based on the graph, the x values are 0°, 45°, 90°, 225°, 360°. Right?
 
  • #13
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Ah, I didn't look carefully. Your Sin[x] curve is not correct.

What is Sin[0]? Sin[45]? Sin[90]? Do you see what's wrong?
 
  • #14
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Ah, I didn't look carefully. Your Sin[x] curve is not correct.

What is Sin[0]? Sin[45]? Sin[90]? Do you see what's wrong?
Do you have any clue of the correct curve for sin x? because that's the only sinx graph that I know.
 
  • #15
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The shape is correct, but you don't have the scaling right. Sin[x] reaches 1 at x = 90°, Sin[x] reaches back to 0 at 180° and so on.
 
  • #16
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The shape is correct, but you don't have the scaling right. Sin[x] reaches 1 at x = 90°, Sin[x] reaches back to 0 at 180° and so on.
Oh yeah I forgot that I have to draw only 1 period. Apologies~ :tongue:
Working on graph..
 
  • #17
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Oh yeah I forgot that I have to draw only 1 period. Apologies~ :tongue:
Working on graph..
No. You have Sin[x] = 1 at x = 45°, Sin[x] = 0 at x = 90°, which is wrong.

You have to draw the curve for 0° ≤ x ≤ 360°.
 
  • #18
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How about this? I hope this one got it right:
2hfncb5.png
 
  • #19
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That's correct. (Though in he graph, you call y = 0 as y = 2).

The red dots are your solution set.
 
  • #20
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:rofl: that should be this:
2llzz7t.png


Sorry about that. I labeled the green line first as y = 1 and accidentally labeled the second one y = 2. What a mess.

So, the solution set is:

x = {0°, 90°, 180°, 360°}

Thank you so much sir! I had fun learning with you. You are good in teaching. I would vote for you as a homework helper here and hope you become one :approve:
 
  • #21
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You're welcome :)

That's very sweet of you, haha
 

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