1. ## general solution qs

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 !

2. Originally Posted by iiharthero
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 !
Use the following:

$\displaystyle \sin A = \sin B \Rightarrow A = B + 2 n \pi$ or $\displaystyle A = (\pi - B) + 2n \pi$ where n is an integer.

$\displaystyle \cos A = \cos B \Rightarrow A = B + 2 n \pi$ or $\displaystyle A = - B + 2n \pi$ where n is an integer.

3. Originally Posted by iiharthero
write general solution of

sin4x - sin2x = 0
$\displaystyle sin4x-sin2x=0$

$\displaystyle 2sin2xcos2x-sin2x=0$

$\displaystyle sin2x(2cos2x-1)=0$

either $\displaystyle sin2x=0$.........(1)

or$\displaystyle cos2x=\frac{1}{2}$..............(2)

for 1st case,
$\displaystyle sin2x=0$

$\displaystyle 2sinxcosx=0$

for 2nd case,
$\displaystyle cos2x=\frac{1}{2}$

$\displaystyle 2cos^2x-1=\frac{1}{2}$

4. Originally Posted by iiharthero
cos3x = cos x
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

5. Originally Posted by yeongil
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
The style of question and the fact that the OP states: "ack, i dont realli (sic) no (sic) wat (sic) sin4x or cos3x or sin3x is"

suggests that the approach advised in my first post is probably the best one for the OP to use.

6. Exactly. If is not necessary to change to "sin(x)" and "cos(x)". To solve "sin(3x)= sin(2x)", just write $\displaystyle 3x= 2x+ 2k\pi$.

7. Originally Posted by HallsofIvy
Exactly. If is not necessary to change to "sin(x)" and "cos(x)". To solve "sin(3x)= sin(2x)", just write $\displaystyle 3x= 2x+ 2k\pi$
and $\displaystyle 3x = (\pi - 2x) + 2k \pi$

where k is an integer.

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# formula of cos^3x

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