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

**idunit14** · November 10th 2010, 12:08 PM

need help solving

**pickslides** · November 10th 2010, 12:29 PM

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

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?

**Prove It** · November 10th 2010, 01:25 PM

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

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

**Prove It** · November 10th 2010, 01:26 PM

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

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

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