1. ## trig problem

Here's the problem:

2cos x + tan x= sec x

I changed the tan x into sin x/cos x

I changed the sec x into 1/cos x

Then I multiplied by cos x and then I factored. I got (2sin x + 3)(3sin x-1)=0

sin x = -3/2 and sin x = 1

But that's not the answer on the answer sheet. What did I do wrong?

2. ok from your first step
$\displaystyle 2cosX + \frac{sinX}{cosX} = \frac{1}{cosX}$

mult thru with $\displaystyle cosX$

$\displaystyle 2cos^2X + sinX = 1$

since

$\displaystyle cos^2X = 1 - sin^2X$

then

$\displaystyle 2(1 - sin^2X) + sinX = 1$

$\displaystyle 2sin^x - sinX - 1 = 0$

$\displaystyle (2sinx + 1)(sinX - 1)$

$\displaystyle sinX = -\frac{1}{2}$ and $\displaystyle sin^{-1}(-\frac{1}{2})$ so x = $\displaystyle -\frac{\pi}{6}$ or $\displaystyle \frac{11\pi}{6}$

3. then what?