Question

2sin²(x)-5cos(x)+1=0

Answer 1

Remember the basic trig identity

<u />$\sin^2 x + \cos^2 x = 1$

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

$x = \frac{\pi}{3}, \frac{5\pi}{3}$