Trig general solution

**srirahulan** · August 31st 2012, 06:27 AM

$cos5x+cosx-sin4x=0$

Do you have any idea to simplify and find a general solution for it.

**Prove It** · August 31st 2012, 07:03 AM

Originally Posted by srirahulan

$cos5x+cosx-sin4x=0$

Do you have any idea to simplify and find a general solution for it.

Try to use angle sum and double angle identities, with the Pythagorean Identity, to reduce this to a polynomial function in sin(x) or cos(x).

**srirahulan** · August 31st 2012, 07:07 AM

I can't understand your solution.

**Prove It** · August 31st 2012, 07:13 AM

Originally Posted by srirahulan

I can't understand your solution.

You could TRY doing what you were instructed. Then I might be able to guide you some more...

**srirahulan** · August 31st 2012, 08:08 AM

$cos5x+cosx-sin4x=0\\2cos3xcos2x-2sin2xcos2x=0\\2cos2x(cos3x-sin2x)=0\\\mbox{Then how can i eliminate sin2x.}$

**Prove It** · August 31st 2012, 08:15 AM

Originally Posted by srirahulan

$cos5x+cosx-sin4x=0\\2cos3xcos2x-2sin2xcos2x=0\\2cos2x(cos3x-sin2x)=0\\\mbox{Then how can i eliminate sin2x.}$

First of all, $\displaystyle \begin{align*} \cos{5x} \end{align*}$ is not $\displaystyle \begin{align*} 2\cos{3x}\cos{2x} \end{align*}$ . If you split it up as $\displaystyle \begin{align*} \cos{(3x + 2x)} \end{align*}$ you will need to use this identity: $\displaystyle \begin{align*} \cos{(\alpha + \beta )} \equiv \cos{\alpha}\cos{\beta} - \sin{\alpha}\sin{\beta} \end{align*}$ . This is the angle sum identity I mentioned in my first post.

It also appears that the $\displaystyle \begin{align*} \cos{x} \end{align*}$ term has disappeared in the second line.

**MaxJasper** · August 31st 2012, 10:54 AM

Originally Posted by srirahulan

$f(x)=cos5x+cosx-sin4x=0$
Do you have any idea to simplify and find a general solution for it.

Let z=x+i y be a complex number. Substitute cos(5x), cos(x), sin(4x) with z component and solve for z...then arguments of z solutions are solutions to f(x) as follows:

$f(z)=\frac{1}{2}\left(z^5+\frac{1}{z^5} \right)+ \frac{1}{2} \left(z+\frac{1}{z}\right)-\frac{1}{2i}\left(z^4-\frac{1}{z^4}\right)$

f(z) simplifies to:

$f(z)=\frac{(-i+z) (i+z) \left(-i+z^2\right) \left(i+z^2\right) \left(1-i z-z^2+i z^3+z^4\right)}{2 z^5} =0$ z is not 0

$z= \left\{-i,i,-(-1)^{1/4},(-1)^{1/4},-(-1)^{3/4},(-1)^{3/4},-\frac{1}{4} i \left(1+\sqrt{5}-i \sqrt{2 \left(5-\sqrt{5}\right)}\right),-\frac{1}{4} i \left(1+\sqrt{5}+i \sqrt{2 \left(5-\sqrt{5}\right)}\right),\frac{1}{4} i \left(-1+\sqrt{5}-i \sqrt{2 \left(5+\sqrt{5}\right)}\right),\frac{1}{4} i \left(-1+\sqrt{5}+i \sqrt{2 \left(5+\sqrt{5}\right)}\right)\right\}$

Arguments of z are as follows:

$x=\left\{-\frac{\pi }{2},\frac{\pi }{2},-\frac{3 \pi }{4},\frac{\pi }{4},-\frac{\pi }{4},\frac{3 \pi }{4},-\pi -\text{ArcTan}\left[\frac{-1-\sqrt{5}}{\sqrt{2 \left(5-\sqrt{5}\right)}}\right],\text{ArcTan}\left[\frac{-1-\sqrt{5}}{\sqrt{2 \left(5-\sqrt{5}\right)}}\right],\text{ArcTan}\left[\frac{-1+\sqrt{5}}{\sqrt{2 \left(5+\sqrt{5}\right)}}\right],\pi -\text{ArcTan}\left[\frac{-1+\sqrt{5}}{\sqrt{2 \left(5+\sqrt{5}\right)}}\right]\right\}$

**BobP** · September 1st 2012, 12:29 AM

$\cos A+\cos B=2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$
so

$\cos 5x+\cos x=2\cos 3x\cos 2x.$

That allows the equation to be written as

$2\cos 2x(\cos 3x - \sin 2x)=0.$

Now use the identities

$\cos 3x = 4\cos^{3}x-3\cos x$ and $\sin 2x = 2\sin x \cos x,$

remove $\cos x$ as a factor and turn what's left inside the brackets into a quadratic in $\sin x.$

**Prove It** · September 1st 2012, 04:56 AM

Originally Posted by srirahulan

$cos5x+cosx-sin4x=0\\2cos3xcos2x-2sin2xcos2x=0\\2cos2x(cos3x-sin2x)=0\\\mbox{Then how can i eliminate sin2x.}$

I apologise for my previous post, you did use a correct identity here.

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