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Trigonometry (1 Viewer)

MrBrightside

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Solve cos3x = 0 for 0 <= x <= pi

I completely forgot how to do these :(

graph shows pi / 2 is where it = 0
 

MrBrightside

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cos^-1(0) = 90 degrees,

1st and 4th quadrant it is positive

goes around 3 times due to the period

90, 2pi - 90, pi + 90, ehh i'm screwed around this part.
 

AAEldar

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cos^-1(0) = 90 degrees,

1st and 4th quadrant it is positive

goes around 3 times due to the period

90, 2pi - 90, pi + 90, ehh i'm screwed around this part.
Firstly, don't use degrees and radians at the same time. Use one or the other and since the question designated radians, use them.

Secondly, is the same as . The last one should be . These are what is equal to, so you then divide these answers by 3.
 

MrBrightside

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oh man, i'm just really bad at this now, I went radians

3x = pi / 2, 2pi - pi / 2, 2pi + pi / 2, 4pi - pi / 2 , now 3rd rev, would it be 3pi + pi / 2, 6pi - pi / 2?
 

AAEldar

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oh man, i'm just really bad at this now, I went radians

3x = pi / 2, 2pi - pi / 2, 2pi + pi / 2, 4pi - pi / 2 , now 3rd rev, would it be 3pi + pi / 2, 6pi - pi / 2?
No no no no!

The original boundary is . The question is . Which means you times the boundaries by 3. So the new boundary is . You only have to go up until around the circle, which is one and a half rotations. Hence the reason why the answers for are only . Divide those answers by 3 and you get .

Make sense?
 
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MrBrightside

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No no no no!

The original boundary is . The question is . Which means you times the boundaries by 3. So the new boundary is . You only have to go up until around the circle, which is one and a half rotations. Hence the reason why the answers for are only . Divide those answers by 3 and you get .

Make sense?
ahh yes much better now, thank you. I dunno why I was getting mixed up with the other method, I was watching maths online vids, but their examples never had a 3 where the x was. I always seem to get thrown by these questions, (use to be so good at them in year 11, then I lost the skill :( )
 

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