Assignment question, root, Intermediate Value Theorem

kuttaman
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Homework Statement



Prove that Prove that the equation has at least one positive real root, using the Intermediate Value Theorem on an appropriate function.

3 arctan(2x-1)=cos^2(x-(∏/6)) + 1

Homework Equations



No clue

The Attempt at a Solution



I honestly speaking have no clue how to even start on this question, I have studied the text but my brain right now is just no functioning. I recently got diagnosed with ADHD and recentlyu started the medication (2days ago) so my head is a little sluggish and fatigued. Math is a bit of a thinking game, and this question is very detrimental for me to get. If someone can answer this/ or help me answer this it would be highly appreciated.
 
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kuttaman said:

Homework Statement



Prove that Prove that the equation has at least one positive real root, using the Intermediate Value Theorem on an appropriate function.

3 arctan(2x-1)=cos^2(x-(∏/6)) + 1

Homework Equations



No clue

The Attempt at a Solution



I honestly speaking have no clue how to even start on this question, I have studied the text but my brain right now is just not functioning. I recently got diagnosed with ADHD and recently started the medication (2days ago) so my head is a little sluggish and fatigued. Math is a bit of a thinking game, and this question is very detrimental for me to get. If someone can answer this/ or help me answer this it would be highly appreciated.
What does the Intermediate Value Theorem state ?
 

Thread 'Finding the nth roots of a complex number'

There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
View full post »

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