Are these two solutions of a trigonometric equation both correct?

Question

$$2\sin^2x+3\cos x=0$$ $$x=(2n+1)\pi \pm \pi/3$$ $$x=2n\pi \pm 2\pi/3$$ I am getting the first one while thinking of solutions lying only in the second and third quadrants and subtracting/adding the required angle from odd multiples of $\pi$. However the second one can easily be arrived at by subtracting/adding the other angle from even multiples of $\pi$. Please let me know where I am going wrong.

Answer 1

Obviously, your two solutions are equivalent: Subtracting $1/3$ from any odd number is the same as adding $2/3$ to any even number and vice versa adding $1/3$ to any odd number is the same as subtracting $2/3$ from any even number.

However, none of these two solutions are correct. If you plug in e.g. $2\pi/3$, you will get $\frac{3}{4} \left( \sqrt{3} - 2 \right)$ but not $0$.

Update

As the OP corrected a typo (see comments), both solutions are correct.

Answer 2

$\{(2n+1)\pi \pm \frac \pi 3|n\in \mathbb Z\} =$

$\{2n\pi + \frac {2\pi}3; (2n+1)\pi + \frac \pi 3| n \in \mathbb Z\}=$

$\{2n\pi + \frac {2\pi}3; (2n + 2)\pi - \frac {2\pi}3| n \in \mathbb Z\}=$

$\{2n\pi + \frac {2\pi}3; 2(n+1)\pi - \frac {2\pi}3|n \in \mathbb Z\} =$

$\{2j\pi + \frac {2\pi}3; 2(k)\pi - \frac {2\pi}3|j= n\text{ for some }n \in \mathbb Z; k = m+1\text{ for some } m \in \mathbb Z\} =$

$\{2n\pi + \frac {2\pi}3; 2n\pi - \frac {2\pi}3|n \in \mathbb Z\}=$

$\{2n\pi \pm \frac {2\pi}3|n \in \mathbb Z\}$.

The two expressions are exactly the same.

(Assuming your method for solving was correct in the first place. I didn't actually attempt to solve.)