Q says solve 2cot^2x -cotx-5=0 in the range -180 to +180. The solutions are -152, -36.51,28.38 and 143.49. I got all except for -152. Some help would be greatly appreciated!

Original post by pigeonwarrior

Q says solve 2cot^2x -cotx-5=0 in the range -180 to +180. The solutions are -152, -36.51,28.38 and 143.49. I got all except for -152. Some help would be greatly appreciated!

Should be 28-180?

Original post by mqb2766

Should be 28-180?

Yup, it is that but I'm confused as to why? As tan inverse when making the equation equal to 0 and solving it for x gives 28.38 and -36.51. Then 180 plus -36.51 gives 143.49. And 180 plus 28.38 will give a solution that does not fall in the given range. Why would they do 28.38-180 for the -152?

Original post by pigeonwarrior

Yup, it is that but I'm confused as to why? As tan inverse when making the equation equal to 0 and solving it for x gives 28.38 and -36.51. Then 180 plus -36.51 gives 143.49. And 180 plus 28.38 will give a solution that does not fall in the given range. Why would they do 28.38-180 for the -152?

You can think about it in terms of tan or cot, and I guess tan is more familiar so a rough working would be

•

The quadratic (in cot) is a "u" and cuts the y axis at -5 so there are two solutions (for cot). As the linear term/gradient is fairly small at that point (-1) the roots will be roughly symmetric about 0.

•

So you have a positive and negative root for tan (1/cot) and therefore the corresponding angles will be roughly equal about 0 so about +/-30. arctan returns values in the range -90..90.

•

Those are the two principle values and tan repeats every 180 so as they want solutions in -180 to 180, then 30 and 30-180 and for the other root gives -30 and -30+180.

•

If there are two roots for the original quadratic, then you will get 4 angles in the range -180..180 as each root will correspond to two angles.

Note if youd used arccot, then the two principle values would be about 30 and 150 as the range of arccot is 0..180. So then the second solution would be obtained by subtracting 180 from each of them.

(edited 7 months ago)

Original post by mqb2766

You can think about it in terms of tan or cot, and I guess tan is more familiar so a rough working would be

Note if youd used arccot, then the two principle values would be about 30 and 150 as the range of arccot is 0..180. So then the second solution would be obtained by subtracting 180 from each of them.

•

The quadratic (in cot) is a "u" and cuts the y axis at -5 so there are two solutions (for cot). As the linear term/gradient is fairly small at that point (-1) the roots will be roughly symmetric about 0.

•

So you have a positive and negative root for tan (1/cot) and therefore the corresponding angles will be roughly equal about 0 so about +/-30. arctan returns values in the range -90..90.

•

Those are the two principle values and tan repeats every 180 so as they want solutions in -180 to 180, then 30 and 30-180 and for the other root gives -30 and -30+180.

•

If there are two roots for the original quadratic, then you will get 4 angles in the range -180..180 as each root will correspond to two angles.

Note if youd used arccot, then the two principle values would be about 30 and 150 as the range of arccot is 0..180. So then the second solution would be obtained by subtracting 180 from each of them.

Thank you so so much, this makes way more sense now!

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