Solve for $\sin^2(x) = 3\cos^2(x)$ I am trying to solve the following equation for x. The textbook's answer and my answer differ and I keep getting the same answer. Could someone please tell me what I'm going wrong? Notice that this is "sin squared x" and 3 * "cos squared x"
$\sin^2x = 3\cos^2x$ //Just rewriting the equation again
$1-\cos^2x = 3\cos^2x$ //Using the Pythagorean identities to substitute in for $\sin^2x$
I then add $\cos^2x$ to both sides yielding:    $$1 = 4\cos^2x$$
I then divide by $4$ yielding:  $$\frac 1 4 = \cos^2x$$
I then take the square root of both sides yielding: $$\frac 1 2 = \cos x$$
Then, I determine that the places where the $\cos x$ is positive $(1/2)$ is $\pi/3$ and $5\pi/3$
The textbook's answer is $\pi/3$ and $2\pi/3$. However the $\cos(2\pi/3)$ is $-(1/2)$ meaning that I must've solved the equation incorrectly.
Does anyone see my mistake?
Thanks!

we know that $\sin^2x=3\cos^2x$ so$$1-\cos^2x=3\cos^2x \Rightarrow 4\cos^2x=1 \Rightarrow \cos^2x=\frac{1}{4} \Rightarrow \cos x=\pm \frac{1}{2}$$
and now we calculate values of $x$
$$\mbox{if} \qquad \cos x=\frac{1}{2},\mbox{ then}\qquad x=\arccos\left(\frac{1}{2}\right)=\frac{\pi}{3},\frac{5\pi}{3}$$
$$\mbox{if} \qquad \cos x=-\frac{1}{2},\mbox{ then}\qquad x=\arccos\left(-\frac{1}{2}\right)=\frac{2\pi}{3},\frac{4\pi}{3}$$
