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

• Nov 10th 2010, 12:08 PM
idunit14
solution in the interval [0,360] cos2theta=square root (3)/2
need help solving
• Nov 10th 2010, 12:29 PM
pickslides
$\displaystyle \displaystyle \cos 2\theta =\frac{\sqrt{3}}{2}$

Make $\displaystyle \displaystyle 2\theta =x$ therefore

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

Using the 30-60-90 special triangle

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

Now what other quadrants is $\displaystyle \cos$ positive in? and how many solutions are we looking for?
• Nov 10th 2010, 01:25 PM
Prove It
Quote:

Originally Posted by idunit14
need help solving

If you want to use a method that does not require as much analysis...

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

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

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

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

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

$\displaystyle \displaystyle \{15^{\circ}, 165^{\circ}, 195^{\circ}, 345^{\circ}\}$.
• Nov 10th 2010, 01:26 PM
Prove It
Quote:

Originally Posted by idunit14
need help solving

If you want to use a method that does not require as much analysis...

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

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

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

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

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

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