Solving Trigonometric Equations

[eutas1]

I don't understand how they got the final answer??
(Refer to attachment A)

 

[Steven G]

n*pi + 13pi/12 = n*pi + 1*pi + pi/12 (since13/12 = 1 + 1/12)
= (n+1)*pi + pi/12.

Since n can be any integer, then n+1 can be any integer. So you can just replace n+1 with m.
So the answer is m*pi + pi/12, where m is an integer.
But that is the same as just saying the answer is n*pi + pi/12.

The n in the last line and the n at the top line are just different n's!

 

[eutas1]

n*pi + 13pi/12 = n*pi + 1*pi + pi/12 (since13/12 = 1 + 1/12)
= (n+1)*pi + pi/12.

Since n can be any integer, then n+1 can be any integer. So you can just replace n+1 with m.
So the answer is m*pi + pi/12, where m is an integer.
But that is the same as just saying the answer is n*pi + pi/12.

The n in the last line and the n at the top line are just different n's!

I still don't understand...

 

[Harry_the_cat]

Your answer is still correct. It is just that their final answer is simpler.

Your solution \(\displaystyle x=\frac{13\pi}{12} + n\pi\) generates the same set of solutions as the simpler solution given.

For your solution,
when \(\displaystyle n=-1\), \(\displaystyle x=\frac{\pi}{12}\),
when \(\displaystyle n=0\), \(\displaystyle x=\frac{13\pi}{12}\),
when \(\displaystyle n=1\), \(\displaystyle x=\frac{25\pi}{12}\), etc.

For the simpler solution,
when \(\displaystyle n=-1\), \(\displaystyle x=\frac{-11\pi}{12}\),
when \(\displaystyle n=0\), \(\displaystyle x=\frac{\pi}{12}\),
when \(\displaystyle n=1\), \(\displaystyle x=\frac{13\pi}{12}\), etc.

 

[lex]

 

[eutas1]

View attachment 26336View attachment 26335

Ah, I get it now! Thank you so much!!

 

[eutas1]

Your answer is still correct. It is just that their final answer is simpler.

Your solution \(\displaystyle x=\frac{13\pi}{12} + n\pi\) generates the same set of solutions as the simpler solution given.

For your solution,
when \(\displaystyle n=-1\), \(\displaystyle x=\frac{\pi}{12}\),
when \(\displaystyle n=0\), \(\displaystyle x=\frac{13\pi}{12}\),
when \(\displaystyle n=1\), \(\displaystyle x=\frac{25\pi}{12}\), etc.

For the simpler solution,
when \(\displaystyle n=-1\), \(\displaystyle x=\frac{-11\pi}{12}\),
when \(\displaystyle n=0\), \(\displaystyle x=\frac{\pi}{12}\),
when \(\displaystyle n=1\), \(\displaystyle x=\frac{13\pi}{12}\), etc.

I still didn't understand after I read your comment, but I get it now after reading the next one. I'm a bit slow... Thank you for your input!!

 

[lex]

Ah, I get it now! Thank you so much!!

View attachment 26336View attachment 26335

Good, as long as you've got it - that's the main thing!
Glad it helped.