Solving this trig equation

[Sophie02]

I have completed part a. But would anybody be able to help me with part b?

 

[Deleted member 4993]

View attachment 26595

I have completed part a. But would anybody be able to help me with part b?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING

Please share your work/thoughts about this problem

 

[Steven G]

You need to look closely at patterns. Pattern recognition is key to mathematics.
I noticed that the four angles in part a were the same.
In part b, with the same formula as in part a, I noticed that three of the four angles were exactly the same. BUT to use this identity in part a ALL the angles must be equal.
What I wrote above is a good hint. Let's see what you can do with it

 

[lex]

Replace the left hand side in part (b) with the right hand side of part (a) (using x instead of [MATH]\theta[/MATH] of course) and then solve.

 

[lex]

For completeness:
[MATH] \begin{align*}\\ \cos x \cot x=\cos x \cot(3x-50)\\ \cos x(\cot x-\cot(3x-50))=0\\ \therefore \cos x=0 \text{ or } \cot x=\cot(3x-50)\\ \\ 0<x<180, \text{ so } \cos x=0 \rightarrow x=90\\ \\ \cot x=\cot(3x-50) \rightarrow 3x-50&=x+180n\\ x&=25+90n\\ 0<x<180 \rightarrow x=25, x=115\\ \\ \text{The solutions are: }x=\text{ 25, 90, 115}\\ \end{align*}\\[/MATH]

 

[Sophie02]

For completeness:
[MATH] \begin{align*}\\ \cos x \cot x=\cos x \cot(3x-50)\\ \cos x(\cot x-\cot(3x-50))=0\\ \therefore \cos x=0 \text{ or } \cot x=\cot(3x-50)\\ \\ 0<x<180, \text{ so } \cos x=0 \rightarrow x=90\\ \\ \cot x=\cot(3x-50) \rightarrow 3x-50&=x+180n\\ x&=25+90n\\ 0<x<180 \rightarrow x=25, x=115\\ \\ \text{The solutions are: }x=\text{ 25, 90, 115}\\ \end{align*}\\[/MATH]

what does the n mean?

 

[Sophie02]

You need to look closely at patterns. Pattern recognition is key to mathematics.
I noticed that the four angles in part a were the same.
In part b, with the same formula as in part a, I noticed that three of the four angles were exactly the same. BUT to use this identity in part a ALL the angles must be equal.
What I wrote above is a good hint. Let's see what you can do with it

Could you show me your working out to see what you think the answer is?

 

[lex]

what does the n mean?

Sorry, [MATH]n[/MATH] stands for any integer.
[MATH]\cot \theta[/MATH] has a period of 180º.
When I got to [MATH]x=25 + 90n[/MATH]I used [MATH]n=0[/MATH] and [MATH]n=1[/MATH] to generate the only two solutions for this part, which are between 0º and 180º.

Looking at the graph of [MATH]\cot x^o[/MATH]
it is clear that [MATH]\hspace2ex \cot x = \cot y \hspace4ex \leftrightarrow \hspace4ex x \text{ and } y[/MATH] differ by a multiple of 180º
i.e. [MATH]x-y = 180n[/MATH], (for some [MATH]n \in \mathbb{Z} \text{), } \hspace2ex \text{ i.e. } x=y+180n[/MATH]
That's how I went from:
[MATH]\cot x=\cot(3x-50) \hspace2ex \text{ to } \hspace2ex 3x-50=x+180n[/MATH]