# Thread: Help with solving a trigometric equation.

1. ## 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 $\left[ 0,4\pi \right]$.

2. 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 $\left[ 0,2\pi \right]$.
$\cos\left(2x+\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}}$

$0 \le x \le 2\pi$

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

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

$2x+ \frac{\pi}{4} = \frac{3\pi}{4} \, , \, \frac{5\pi}{4} \, , \, \frac{11\pi}{4} \, , \, \frac{13\pi}{4}$

3. Thanks, kinda made it clearer.

4. Nice solution skeeter, the graph checks it.

x1 = (pi/2)(2k - 1), k is an element of Z

x2 = (pi/4)(4k + 1), k is an element of Z

5. So I think my mistake would be that I should have treated the entire equation and find all the angles within the domain first before solving for x. What I did was solve for x then add/minus the 2pi to get the other angles, which was wrong.