trigonometry equation, $[\cos(2x)]^2-\sin(2x)=1$ I tried to solve this equation: $[\cos(2x)]^2-\sin(2x)=1$
I got different answer from the book.
the answer in the book: $x=-45+180k$ , $x=90k$
Am I right? $x = -45+180k$ is equal to $x= 135+180k$ ? How can I check if it's the same? Or maybe I did a mistake? please help
\begin{align*}
[\cos(2x)]^2-\sin(2x) & = 1\\
-\sin(2x)& = 1-\cos(2x)^2\\
-\sin(2x)& = \sin(2x)^2\\
-\sin(2x)-\sin(2x)^2 & = 0\\
\sin(2x)+\sin(2x)^2 & = 0\\
\sin(2x)[1+\sin(2x)] & = 0
\end{align*}
\begin{align*}
\sin(2x) & =0 & 1 + \sin(2x) & = 0\\
2x & = 180k & \sin(2x) & = -1\\
x & = 90k &  2x & = 270+360k\\
& & x & = 135+180k
\end{align*}

$$\cos²(2x)-\sin(2x) = 1$$
$$-\sin(2x)= 1-\cos²(2x)$$
$$-\sin(2x)=\sin²(2x)$$
$$0= \sin²(2x)+\sin(2x)$$
$$\sin(2x)(\sin(2x)+1)=0$$
Then, you get two equations:
$$\sin(2x)=0$$
and 
$$\sin(2x)+1=0$$
$$\sin(2x)=\sin (180)$$
and 
$$\sin(2x)=\sin(270)$$
From the $$\sin(2x)=\sin(180)$$, you can get the solution:
$$2x=180+k.360$$ or $$x=90+k.180$$ 
and
$$2x= k.360$$ or $$x= k.180$$
From the $$\sin(2x)=\sin(270)$$, you can get the solution:
$$2x=270+k.360$$ or $$x=135+k.180$$
and 
$$2x=(180-270)+k.360$$ or $$x= -45+k.180$$
So, the solution:
$x=k.180$ 
$x=90+k.180$
$x=135+k.180$
$x=-45+k.180$
