Is this correct about csc function?

Question

the question says :

solve $2\csc^2x+3\csc x+1=0$ over the interval $[0,2\pi)$.

I tried to solve this question and i found out that there is only one solution which is $3\pi/2$ ..

is this correct?

Answer 1

Yes. It's correct. The quadratic in csc(x) has two solutions, but one of them is inadmissible (unless x is allowed to be complex). The only value that gives a valid solution is csc(x) = -1, giving you the sole root x = 3pi/2.

Answer 2

Let $a = csc(x)$, then you have: $2a^2 + 3a + 1 = 0$

Doing the quadratic formula you get:

$a = {-3 \pm 1\over 4}$ or $a=-1, a =-1/2$

Now, $-1 = a = csc(x)$, so $-1 = {1 \over sin(x)}$ or $sin(x) = -1$. So $x = {3\pi\over2}$.

Solving for the other equation we get $-1/2 = csc(x)$ so $-2 = sin(x)$, which can never happen because $-1 <= sin(x) <= 1$. Thus $3\pi \over 2$ is the only answer.