# Trig Solve for Solution Within Interval

1. Sep 28, 2006

### catenn

Hi, I have a problem that says to solve this equation for the interval 0<theta<2 pie which is one revolution of the unit circle. The equation is:

(square root of 2)(cos 2theta) = 1

The cos2theta is a double angle, but I couldn't find any solutions for theta that would = 1 at the end. The answer has to fit in for the theta angle. Is this possibly no solution? I had originally put pie/4 as an answer but realized I forgot the parenthesis in the calculator and now I need to redo the problem. Thanks!

2. Sep 28, 2006

### Data

Draw the unit circle in the x-y cartesian plane. If you rotate around the circle by an angle $\theta$ counterclockwise wrt the positive x-axis, then $\cos{\theta}$ is the x-coordinate of the resulting point. So what $\theta$ will make $\cos{\theta} = \frac{1}{\sqrt{2}}$ (HINT: What's $2(1/\sqrt{2})^2$?)?

Edit: courtrigrad I suggest plugging $\pi / 2$ into $\sqrt{2}\cos{2\theta}$ and seeing what you get :tongue2:

Last edited: Sep 28, 2006
3. Sep 28, 2006

my fault it should have been $$\theta = \frac{\pi}{8}$$. I multiplied instead of divided.

4. Sep 28, 2006

### Data

You can give a closed form for $\theta$.

Edit:

5. Sep 28, 2006

$$\theta = \frac{\pi}{8} + 2n\pi$$

6. Sep 28, 2006

### Data

he said $0 < \theta < 2\pi$ though (of course, there's still $\theta = \frac{15}{8} \pi$!) :tongue:

Last edited: Sep 28, 2006
7. Sep 28, 2006

### catenn

Thank you both so much for the help! I really appreciate it and understand now.