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tikki14 · Apr 21, 2011, 04:48 AM

I meet some problems in solving some equations. I tried a lot of methods, but I didn't get any solution, so I'd be grateful if you could give some ideas.

1) $\sin^3 x + \cos^3 x = \cos 2x$ , for $-90 \circ < x < 90 \circ$

At first, I tried to write

$\sin^3 x + \cos^3 x = (\sin x + \cos x)(\sin^2 x + \cos^2 x - \sin x \cos x)$
$= (\sin x + \cos x)(1- \sin x \cos x)$

but I don't know how to solve next.

Then, I wrote $\cos 2x = \cos^ x - \sin^2 x \Rightarrow \sin^3 x + \cos^3 x - \cos^2 x + \sin^2 x= 0$

If $\cos x = 0 \Rightarrow \sin^3 x + \sin^2 x = 0 --> \sin^2 x (\sin x + 1) = 0$
$\sin^2 x = 0 \Rightarrow x = \arctsin 0 + k \pi = k \pi, k \epsilon Z$
$sin x + 1 = 0 \Rightarrow x = (-1)^(k+1) arcsin 1 + k \pi = (-1)^(k+1) \frac{\pi}{2} + k \pi, k \epsilon Z$

If \cos x \neq 0 \Rightarrow \frac{\sin^3 x}{\cos^3 x} + \frac{\cos^3 }{\cos^3 x} - \frac{\cos^2 x}{\cos^3 x} + \frac{\sin^2 x}{\cos^3 x} = 0 \Rightarrow \tan^3 x + 1 - \frac{1}{\cos x} + \tan^2 \frac{1}{cos x} = 0 \Rightarrow \tan^3 x + 1 - \frac{1}{cos x} (\tan^2 x - 1) = 0 \Rightarrow (\tan x + 1)(\tan^2 x - \tan x + 1) - \frac{1}{\cos x} (\tan x -1)(\tan x + 1) = 0 \Rightarrow (\tan x +1)[\tan^2 x - \tan x +1 - \frac{1}{\cos x}(\tan x -1)] = 0 \Rightarrow (\tan x +1)[\tan x (\tan x -1) + 1 - \frac{1}{\cos x}(\tan x - 1)] = 0 \Rightarrow (\tan x +1)[(\tan x -1)(\tan x - \frac{1}{\cos x})+1] =0 \Rightarrow (\tan^2 -1)(\tan x - \frac{1}{\cos x} + \tan x + 1) = 0 \Rightarrow (\tan^x -1)(2 \tan x - frac{1}{\cos x} + 1) = 0[/math]

$\tan^2 x -1 = 0 \Rightarrow \tan x = \pm 1$
$\tan x = 1 \Rightarrow x= \frac{\pi}{4} + k \pi, k \in Z$
$\tan x = -1 \Rightarrow x = - \frac{\pi}{4} + k \pi, k \in Z$
$2 \tan x - \frac{1}{\cos x} + 1 = 0 \Rightarrow 2 \tan x + 1 = \frac{1}{sqrt(1 + \tan^2 x)} \Rightarrow (2 \tan x + 1)^2 = \frac{1}{1 + \tan^2 x} \Rightarrow (1 + \tan^2 x)(2 \tan x +1)^2 = 1$

Is this correct? If it is, how to solve next?

2) $\cot x - \tan x + 2 (\frac{1}{\tan x + 1} + \frac{1}{\tan x -1}) = 4$

$\frac{1}{\tan x} - \tan x + 2 \frac{\tan x -1 + \tan x + 1}{\tan^2 - 1} = 4$

$\frac{1 - \tan^2 x}{\tan x} + \frac{4 \tan x}{\tan^2 x -1} = 4$

$-(\tan^2 x -1)^2 + 4 \tan^2 x - 4 \tan x (\tan^2 x -1) = 0$

What to do now to get the solution?

Thank you for taking time to read this.

ebaines · Apr 21, 2011, 06:12 AM

Try this forteh right hand side of the original equation:
$ \cos (2x) = \cos ^2x - \sin ^2x = (\cos x - \sin x)(\cos x + \sin x) $

You already have the left hand side as
$ (\sin x + \cos x)(\sin ^2 x + \cos ^2 x - \sin x \cos x) $

So you can divide through by $\sin x + \cos x$ ; however, you can do this only if $\sin x + \cos x \ne 0$ . Let's hold that thought for later:

$ \sin ^2 x + \ cos^2 x - \sin x \cos x = \cos x - \sin x \\ sin^2 x + (\cos x - \sin x) (cos x -1)= 0\\ 1 - cos^2 x = ( \cos x - \sin x )(1 - \cos x)$

Now you can divide through by $(1 - \cos x )$ , but only if $(1 - \cos x ) \ne 0$ :

$ 1 + \cos x = \cos x - \sin x\\ 1 = - \sin x\\ x = - \pi /2 $

So that's one solution. Now go back and consider the cases that we threw out earlier; namely

$ \sin x + \cos x = 0 $
and
$ 1 - \cos x = 0 $

This gives you two more solutions, which I'll leave to you to solve.

tikki14 · Apr 21, 2011, 02:57 PM

Thank you ebaines! It is easier than thought. I should have kept [math]sin^2 x + cos^2 x[math] on the left side instead of writing 1.