Question

solve:

cos^(2x)-sin(2x)=0

Answer 1

please clarify cos^(2x) and please give me a medal for the last problem :P

Answer 2

my b,

cos^2(2x)-sin^2(2x)=0

Answer 3

\[2x=\theta \]
\[\cos^2(\theta)-\sin^2(\theta)=0 \iff \cos^2(\theta)=\sin^2(\theta)\]
\[\rightarrow \cos(\theta)=\pm \sin(\theta)\]
\[\theta=\frac{k \pi}{4},k \in \mathbb{Z} \iff 2x=\frac{k \pi}{4}, k \in \mathbb{Z} \iff x=\frac{k \pi}{8}, k \in \mathbb{Z}\]

Answer 4

x=2npi-pi/4
n is an integer

Answer 5

\[\forall n \in \mathbb{Z}, 2 \pi n=0[\mod2\pi]\]
So there's no need for the 2npi