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

Since $\sin^2 x=1-\cos^2 x$ the given equation is equivalent to
\begin{align}
3\cos^2 x&=1-\cos^2 x\\
\iff \quad \cos^2 x&=\frac{1}{4}\\
\iff \cos x&\in\left\{-\frac{1}{2},\frac{1}{2}\right\}
\end{align}
Hence $$x=180^{\circ}\cdot k \pm 60^{\circ}$$
where $k$ is an integer number.

Silas2033: Notice that in this problem $\cos x$ can be either $\frac{1}{2}$ or $-\frac{1}{2}$, then you must regard two cases:
1) $\cos x = \cos y$ giving us $x=\pm y +360^{\circ}k=\pm y +180^{\circ}(2n)$ and 
2) $\cos x = -\cos y$ giving us $x=180^{\circ}-(\pm y +360^{\circ}k)=\mp y - 180^{\circ}(2k-1)$.
