Trig. Finding the general solution

**akpolish** · December 16th 2008, 04:14 PM

2 cos x = sec x

Thank you ahead of time.

**chabmgph** · December 16th 2008, 04:23 PM

Originally Posted by akpolish

2 cos x = sec x

Thank you ahead of time.

Hint: $\sec x = \frac{1}{\cos x}$

Try to solve for $\cos x$ and see.

**akpolish** · December 16th 2008, 04:25 PM

Thank you for the hint but I still don't understand.

**chabmgph** · December 17th 2008, 03:36 PM

Originally Posted by akpolish

2 cos x = sec x

Thank you ahead of time.

Originally Posted by akpolish

Thank you for the hint but I still don't understand.

$2 \cos x = \sec x$
$2 \cos x =\frac{1}{\cos x} \text{ \; \; \; \; (}\cos x \neq 0$ )
$2 \cos^2 x =1$
$\cos^2 x =\frac{1}{2}$
$\cos x =\pm\frac{\sqrt{2}}{2}$

Now see if you can find the angles which gives you these value.

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