# Thread: Solving a trigonometric equation.

1. ## Solving a trigonometric equation.

I was asked to solve the following equation:

$\displaystyle 2sin(x) + csc(x) = 3$

I solved it by changing the csc(x) to $\displaystyle \frac{1}{sin(x)}$ and then by multiplying through by sin(x), yielding:

$\displaystyle 2sin^2(x) - 3sin(x) + 1 = 0$

I then used the "completing the square" method to solve the equation and arrived at sin(x) = 1. Because I ended up with such a neat answer, I'm wondering if I missed some easier way to solve the problem.

Was there a more simple way I could have solved this that wouldn't have involved completing the square or using the quadratic equation?

Thanks!

2. There isn't a simpler method of solution. However, there is another solution to your quadratic. (The solution you found is one solution.)

3. Oh, did you mean sin(x) = +/- 1?

Thanks!

4. Oh, sorry. It should be sin(x) = 1 and sin(x) = 1/2, correct?

5. Alas, no. If you plug in $\displaystyle \sin(x) = -1$, it will not satisfy the equation. You know that $\displaystyle \sin(x) = 1$ is a solution. What do you need to multiply $\displaystyle (\sin(x) - 1)$ by to get $\displaystyle 2\sin^{2}(x)-3\sin(x)+1?$

[EDIT] Ja, you got it.

6. So, now what? For what range of x are you looking for solutions?

7. $\displaystyle 0 < or = X < 2 \pi$

So:

$\displaystyle \frac{\pi}{2}, \frac{\pi}{6}, \frac{5\pi}{6}$

8. I'd say you're correct.