# Math Help - solution in the interval [0,360] cos2theta=square root (3)/2

1. ## solution in the interval [0,360] cos2theta=square root (3)/2

need help solving

2. $\displaystyle \cos 2\theta =\frac{\sqrt{3}}{2}$

Make $\displaystyle 2\theta =x$ therefore

$\displaystyle \cos x =\frac{\sqrt{3}}{2}$

Using the 30-60-90 special triangle

$\displaystyle x =\frac{\pi}{6}\implies 2\theta = \frac{\pi}{6} \implies \theta = \frac{\pi}{12}$

Now what other quadrants is $\cos$ positive in? and how many solutions are we looking for?

3. Originally Posted by idunit14
need help solving
If you want to use a method that does not require as much analysis...

$\displaystyle \cos{2\theta} = \frac{\sqrt{3}}{2}$

$\displaystyle 2\theta = \{30^{\circ}, 360^{\circ} - 30^{\circ}\} + 360^{\circ}n$ where $\displaystyle n$ is an integer

$\displaystyle 2\theta = \{30^{\circ}, 330^{\circ}\} + 360^{\circ}n$

$\displaystyle \theta = \{15^{\circ}, 165^{\circ}\} + 180^{\circ}n$.

So the solutions in the domain $\displaystyle [0, 360]$ are

$\displaystyle \{15^{\circ}, 165^{\circ}, 195^{\circ}, 345^{\circ}\}$.

4. Originally Posted by idunit14
need help solving
If you want to use a method that does not require as much analysis...

$\displaystyle \cos{2\theta} = \frac{\sqrt{3}}{2}$

$\displaystyle 2\theta = \{30^{\circ}, 360^{\circ} - 30^{\circ}\} + 360^{\circ}n$ where $\displaystyle n$ is an integer

$\displaystyle 2\theta = \{30^{\circ}, 330^{\circ}\} + 360^{\circ}n$

$\displaystyle \theta = \{15^{\circ}, 165^{\circ}\} + 180^{\circ}n$.

So the solutions in the domain $\displaystyle [0, 360]$ are

$\displaystyle \{15^{\circ}, 165^{\circ}, 195^{\circ}, 345^{\circ}\}$.