Trigonometry

[Loki123]

How do I get the actual solution?

 

[blamocur]

While you are technically correct you can get to the mentioned answer by using $u=4x$, then $2\sin u = \sqrt{2-\sqrt{3}}$, square both parts and use some trigonometric identities. Can you work out the rest?

 

[Loki123]

While you are technically correct you can get to the mentioned answer by using $u=4x$, then $2\sin u = \sqrt{2-\sqrt{3}}$, square both parts and use some trigonometric identities. Can you work out the rest?

I seem to be going in circles.

 

[Loki123]

While you are technically correct you can get to the mentioned answer by using $u=4x$, then $2\sin u = \sqrt{2-\sqrt{3}}$, square both parts and use some trigonometric identities. Can you work out the rest?

I got it, kinda. The only difference is I got kPi/4 instead of kPi/2 and 5Pi/48 instead of 11Pi/48

 

[Loki123]

I got it, kinda. The only difference is I got kPi/4 instead of kPi/2 and 5Pi/48 instead of 11Pi/48View attachment 32943
View attachment 32945

I made a mistake, I know why it's supposed to be 11Pi, but what about kPi/4 instead of kPi/2

 

[blamocur]

I made a mistake, I know why it's supposed to be 11Pi, but what about kPi/4 instead of kPi/2

I've made the same mistake of getting the period of $\frac{k\pi}{4}$ instead of $\frac{k\pi}{2}$. The error arose when we squared the equation :
$4\sin^2 u = 2-\sqrt{3}$ is not equivalent to $2\sin u = \sqrt{2-\sqrt{3}}$, but it is equivalent to $\pm \left( \sqrt {2-\sqrt{3}}\right)$. I.e., by squaring we've introduce additional "solutions" (there is a term for them, but I don't remember what it is).

 

[Deleted member 4993]

(there is a term for them, but I don't remember what it is).

Extraneous solution!!

 

[blamocur]

Extraneous solution!!

Thank you!

 

[Steven G]

Extraneous solution!!

I call it corner time.

 

[Deleted member 4993]

I call it corner time.

Are you feeling lonely there - I am sitting here in the other corner - otis sent me here to keep you company....