+505 Find solutions with \(x\in [0,2\pi]\) to the equation \(5\sec(x)=3\tan^2(x)+1 . \)
Okay I had an idea for this question:
First, we turn the 3 sec(x) into 3/cos(x). After that, turn 3tan^2(x) into (3 * 1-cos(x))/ 1 + cos(x). I don't know where to go from here. Help needed :D
+876 $\sec(x) = \frac{1}{\cos(x)}$
\(\begin{align*} \tan^2(x) &= \left(\frac{\sin x}{\cos x} \right)^2 \\ &= \frac{\sin^2 x}{\cos^2 x} \\ &= \frac{\sin^2 x}{\cos^2 x} + \frac{\cos^2 x}{\cos^2 x} - 1\\ &= \frac{\sin^2 x + \cos^2 x}{\cos^2 x} - 1\\ &= \frac{1}{\cos^2 x} - 1 \end{align*}\)
$\frac{5}{\cos(x)} = \frac{3}{\cos^2 x} - 2$
$5 \cos(x) = 3 - 2 \cos^2 x$
$2 \cos^2 x + 5 \cos x - 3 = 0$
$(\cos x + 3) (2 \cos x - 1) = 0$
$\cos x = -3$
$\cos x = \frac{1}{2}$
I think you can take it from here.
+505 I'm still a bit confused. So I would then know that one of the answers in pi/3. But I don't know what other answers there are...
+876 Hi Mobius,
You are correct. There is a second answer to the question, but does it fit in the range?
Best,
MPS
