# Help with solving a trigometric equation.

• Sep 11th 2009, 06:32 PM
n0d3
Help with solving a trigometric equation.
Hello,

The equation is as follows:
$\cos 2\left( x_{+\frac{\pi }{8}} \right)=-\left( \frac{1}{\sqrt{2}} \right)$

With the interval $\left[ 0,2\pi \right]$.

So what I did was let $2x+\frac{\pi }{4}=\frac{3\pi }{4}$ and $2x+\frac{\pi }{4}=\frac{5\pi }{4}$ since cos is negative in the 2nd and 3rd quadrants with a basic angle of $\frac{\pi}{4}$.

I then continued to solve the two equations and ended up with x= $\frac{\pi }{4}$ and $\frac{\pi }{2}$.

However I'm missing out the two angles $\frac{5\pi }{4}$ and $\frac{3\pi }{2}$. I'm supposed to be adding $2\pi$ to $\frac{\pi }{4}$ and $\frac{\pi }{2}$ right? But it doesn't give me the other two angles that I missed out.

Also, regarding intervals, when I multiply the a number into the brackets, does it change? Like for the above when I multiply the 2 to get 2x, does the interval get multiplied by two also? That is $\left[ 0,4\pi \right]$.
• Sep 11th 2009, 06:44 PM
skeeter
Quote:

Originally Posted by n0d3
Hello,

The equation is as follows:
$\cos 2\left( x_{+\frac{\pi }{8}} \right)=-\left( \frac{1}{\sqrt{2}} \right)$

With the interval $\left[ 0,2\pi \right]$.

$\cos\left(2x+\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}}$

$0 \le x \le 2\pi$

$0 \le 2x \le 4\pi$

$\frac{\pi}{4} \le 2x + \frac{\pi}{4} \le \frac{17\pi}{4}$

$2x+ \frac{\pi}{4} = \frac{3\pi}{4} \, , \, \frac{5\pi}{4} \, , \, \frac{11\pi}{4} \, , \, \frac{13\pi}{4}$
• Sep 11th 2009, 07:07 PM
n0d3