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!
Best Answer
You are correct until the square root: This leads to
$$\cos{x} = \pm \frac 1 2$$