Question

explain
2cos^2(4x)-1=0

Answer 1

\(\cos(8x) = 0\)

\(8x = \pi/2 + 2k\pi, ~~~~~3\pi/2 + 2n\pi\)

Answer 2

\[8x=\frac{ \pi }{ 2 }+n \pi\]
where n is an integer.

Answer 3

Alternatively,
\[2\cos^2(4x)-1=\left(\sqrt2\cos(4x)-1\right)\left(\sqrt2\cos(4x)+1\right)=0\]giving you two equations to solve,
\[\begin{cases}\sqrt2\cos(4x)-1=0\\[1ex]\sqrt2\cos(4x)+1=0\end{cases}\]More work than necessary, but it doesn't hurt :P

Answer 4

U know
Cos(2A)=2*cos^2A-1

Answer 5

So plz tell me
Cos (8A)=?