Remember the basic trig identity
\(\sin^2 x + \cos^2 x = 1\)
We can move the cosine term to the right side to get
\(\sin^2 x = 1 - \cos^2 x\)
We can then replace \(\sin^2 x\) in the original equation and basically factor the resulting equation like a quadratic:
\(2(1 - \cos^2 x) - 5 \cos x + 1 = 0\)
\((2 - 2 \cos^2 x) - 5 \cos x + 1 = 0\)
\({-2} \cos^2 x - 5 \cos x + 3 = 0\)
\(2 \cos^2 x + 5 \cos x - 3 = 0\)
\((2 \cos x - 1)(\cos x + 3) = 0\)
Then, we can see that \(\cos x\) either equals \(\frac{1}{2}\) or \(-3\). But since the range of cosine is only from \([-1, 1]\), cosine can't equal \(-3\) at all! So, you just have to solve for x when \(\cos x = \frac{1}{2}\), which is when
\(x = \{ \frac{\pi}{3}, \frac{5\pi}{3} \} + 2 \pi k | k \in \mathbb{Z}\)
(The last part of the solution says that k can be any integer.)
If you only want solutions in the range \([0, 2\pi)\), then your answer is just
\(x = \frac{\pi}{3}, \frac{5\pi}{3}\)
