<s>[INST] trigonometry equation $3\cos(x)^2 = \sin(x)^2$ I tried to solve this equation, but my solution is wrong and I don't understand why. the answer in the book is: $x = \pm60+180k$. my answer is: $x= \pm60+360k$.
please help :)
3cos(x)^2 = sin(x)^2
3cos(x)^2 = 1 - cos(x)^2

t = cos(x)^2

3t=1-t
4t=1
t=1/4

cos(x)^2 = 1/4
cos(x) = 1/2
cos(x) = cos(60)

x = +-60+ 360k

edited:
cos(x) = -1/5
cos(x) = cos(120)
x = +-120 + 360k

I still don't get the answer in the book
 [/INST] \begin{align*}
3\cos^2 x & = \sin^2x\\
3 & = \frac{\sin^2x}{\cos^2x}\\
\tan^2x & = 3\\
\tan x & = \sqrt{3}\\
\tan x & = \tan 60
\end{align*}
From there, you can use the tangent equation:
$$x= \pm 60+180k$$
</s>