solve for x: 4cos 2x + 3
- Thread starter xomandi
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- Apr 12, 2005
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There are some very difficult notation problems on this one.
First, solve what? You have not provided an equation. From your second step, I see that you intended an equation. It is not the same if you fail to write "=0". That's what makes it an equation.
Second, you seem to be on the right track, but please think about what you are writing. Here is what you have written: \(\displaystyle 1\;+\;\cos^{2}{\frac{x}{2}}\). Is that what you mean? I believe it is not. Perhaps you meant: \(\displaystyle 1\;+\;\frac{\cos^{2}{x}}{2}\)? Maybe \(\displaystyle \frac{1\;+\;\cos^{2}{x}}{2}\). If you did mean one of those, what is it for? Again, you have not provided an equation.
Since \(\displaystyle \cos{2x}\;=\;2\cos^{2}{x}\;-\;1\), we have \(\displaystyle \cos{x}\;=\;\sqrt{\frac{(\cos{2x})+1}{2}}\) without consideration of the Quadrant of either.
Okay, where does that leave us?
Note: PLEASE be more careful with your notation. Make sure what you write is what you mean.
First, solve what? You have not provided an equation. From your second step, I see that you intended an equation. It is not the same if you fail to write "=0". That's what makes it an equation.
Second, you seem to be on the right track, but please think about what you are writing. Here is what you have written: \(\displaystyle 1\;+\;\cos^{2}{\frac{x}{2}}\). Is that what you mean? I believe it is not. Perhaps you meant: \(\displaystyle 1\;+\;\frac{\cos^{2}{x}}{2}\)? Maybe \(\displaystyle \frac{1\;+\;\cos^{2}{x}}{2}\). If you did mean one of those, what is it for? Again, you have not provided an equation.
Since \(\displaystyle \cos{2x}\;=\;2\cos^{2}{x}\;-\;1\), we have \(\displaystyle \cos{x}\;=\;\sqrt{\frac{(\cos{2x})+1}{2}}\) without consideration of the Quadrant of either.
Okay, where does that leave us?
Note: PLEASE be more careful with your notation. Make sure what you write is what you mean.