Trigonometric equation

[Kondwani Hauya]

Find all the general solutions of the trigonometric equation 3 sin 2x = - 1
Solution.
sin 2x = -1/3
I can't finalize, may somebody help please!

 

[MarkFL]

I would be inclined, as a next step, to write:

[MATH]2x=\frac{3\pi}{2}\pm\arccos\left(\frac{1}{3}\right)+2\pi k[/MATH] where \(k\in\mathbb{Z}\)

Do you see why this works, and can you proceed?

 

[Kondwani Hauya]

[MATH]{2x}={arc cos (-1/3)+π+2kπ}frac{2}[/MATH]

 

[Harry_the_cat]

Shoudn't it be arcsin rather than arccos ??

\(\displaystyle 2x = k\pi -arcsin(\frac{1}{3})\)

EDIT: This in incorrect see below. Sorry.

 

[Kondwani Hauya]

May you show how?

 

[Harry_the_cat]

Sorry my answer in Post #4 is incorrect.

Draw a diagram. Which two quadrants must 2x lie in, ie in which 2 quadrants is sin(2x) negative?

Solution:
\(\displaystyle 2x = k\pi - arcsin(\frac{1}{3})\) if |k| is even or 0
or
\(\displaystyle 2x = k\pi + arcsin(\frac{1}{3})\) if |k| is odd

 

[Kondwani Hauya]

Is it possible to find the value of x by dividing both numerators by 2?

 

[Harry_the_cat]

What do you mean by both numerators?

You can find x by dividing both sides by 2.

 

[Kondwani Hauya]

I was trying to say dividing both sides by 2 like this:
[X]={/kπarcsin(1/frac3)}/frac{2}

 

[MarkFL]

Shoudn't it be arcsin rather than arccos ??

\(\displaystyle 2x = k\pi -arcsin(\frac{1}{3})\)

EDIT: This in incorrect see below. Sorry.

I chose to use the inverse cosine function along with the symmetry of the solution about the angle [MATH]\frac{3\pi}{2}[/MATH] to state the solution more succinctly.