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
 A: \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:
$$x= \pm 60+180k$$
A: 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 $x=180^{\circ}-(\pm y +360^{\circ}k)=\mp y - 180^{\circ}(2k-1)$.
A: Using $\cos2A=1-2\sin^2A=2\cos^2A-1,$
$$3\cdot\dfrac{1+\cos2A}2=\dfrac{1-\cos2x}2\iff\cos2x=-\dfrac12=-\cos60^\circ$$
and as $\cos(180^\circ-y)=-\cos y$
$$\cos2x=\cos(180^\circ-60^\circ)\iff2x=360^\circ n\pm120^\circ$$ where $n$ is any integer
