write general solution of
sin4x - sin2x = 0
cos3x = cos x
sin3x = sin2x
ack, i dont realli no wat sin4x or cos3x or sin3x is, or how to get to it, so if anybody could help me, i would greatly appreciate it
thanx !
$\displaystyle sin4x-sin2x=0$
$\displaystyle 2sin2xcos2x-sin2x=0$
$\displaystyle sin2x(2cos2x-1)=0$
either $\displaystyle sin2x=0$.........(1)
or$\displaystyle cos2x=\frac{1}{2}$..............(2)
for 1st case,
$\displaystyle sin2x=0$
$\displaystyle 2sinxcosx=0$
for 2nd case,
$\displaystyle cos2x=\frac{1}{2}$
$\displaystyle 2cos^2x-1=\frac{1}{2}$
There is a little-known formula for cos 3x:
$\displaystyle \cos 3x = 4\cos^3 x - 3\cos x$
Here is how it's derived:
$\displaystyle \begin{aligned}
\cos 3x &= \cos (2x + x) \\
&= \cos 2x \cos x - \sin 2x \sin x \\
&= (2\cos^2 x - 1)\cos x - (2\sin x \cos x)\sin x \\
&= 2\cos^3 x - \cos x - 2\cos x \sin^2 x \\
&= 2\cos^3 x - \cos x - 2\cos x(1 - \cos^2 x) \\
&= 2\cos^3 x - \cos x - 2\cos x + 2\cos^3 x \\
&= 4\cos^3 x - 3\cos x
\end{aligned}$
Back to the problem:
$\displaystyle \begin{aligned}
\cos 3x &= \cos x \\
4\cos^3 x - 3\cos x &= \cos x \\
4\cos^3 x - 4\cos x &= 0 \\
4\cos x(\cos^2 x - 1) &= 0 \\
4\cos x(\cos x + 1)(\cos x - 1) &= 0
\end{aligned}$
Now set each factor equal to 0 and solve for x...
01