0
$\begingroup$

$4\cos^2 \left( x + \dfrac{1}{4}\pi \right)$ = 3

My final answer:

$ x = \frac{11}{12}\pi+k\pi $ and $x = \frac{7}{12}\pi + k\pi $

In the correction model it is $x = \frac{7}{12}\pi + k\pi $ and $x = -\frac{1}{12}\pi+k\pi$ (and $x = -\frac{1}{12}\pi+k\pi$ equals $x = 1\frac{11}{12}\pi+k\pi$ and not $ x = \frac{11}{12}\pi+k\pi $

  • 0
    I usually have trouble i$n$terpreti$n$g a$n$swers like you gave me, I am just puzzled a$n$d if you read my reaction to siminore below you will understand why I'm extremely puzzled2012-09-15

1 Answers 1

1

Since $ -\frac{1}{12}=\frac{11}{12}-1 $ you found exactly the same solutions.

  • 0
    It has nothing to do with integers. $\cos(x+\pi)=-\cos x$; if you can prove that, then squaring both sides you get $\cos^2(x+\pi)=\cos^2x$, so square of cosine has period $\pi$. By the way, if you want to be sure I see a comment intended for me, you have to include @Gerry in it somewhere.2012-09-16