Thread: Just want to verify I am doing this right

1. Just want to verify I am doing this right

I did get the right answers, but I still would like to make sure my thinking is accurate, and also check if I made unnecessary calculations somewhere.

Get all the solutions to the trigonometric equation $\sin^4 x + \cos^4 x = 1$ in the interval [0, 2pi[

What I did was:

$\sin x\sin x\sin x\sin x + \cos x\cos x\cos x\cos x = 1$

$
[-\dfrac{1}{2}(\cos 2x - \cos 0)][-\dfrac{1}{2}(\cos 2x - \cos 0)] + [\dfrac{1}{2}(\cos 2x + \cos 0)][\dfrac{1}{2}(\cos 2x + \cos 0)] = 1$

$
[-\dfrac{1}{2}(2\cos^2 x - 2)][-\dfrac{1}{2}(2\cos^2 x - 2)] + [\dfrac{1}{2}(2\cos^2 x)][\dfrac{1}{2}(2\cos^2 x)] = 1$

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

$2\cos^4 x - 2\cos^2 x + 1 = 1$

$\cos^4 x - \cos^2 = 0$

We set u = cos^2 and get $u^2 - u = 0$ which gives us u = 0 and u = 1.

cos^2 x = 0 gives us cos x = +/- 0. x = pi/2 + k(pi).
cos^2 x = 1 gives us cos x = +/- 1. x = 0 + k(pi).

So the solutions are, in order,
0, pi/2, pi, 3pi/2.

I apologize for the mix and match between Latex and regular text.

2. Originally Posted by Sabo
I did get the right answers, but I still would like to make sure my thinking is accurate, and also check if I made unnecessary calculations somewhere.

Get all the solutions to the trigonometric equation $\sin^4 x + \cos^4 x = 1$ in the interval [0, 2pi[

What I did was:

$\sin x\sin x\sin x\sin x + \cos x\cos x\cos x\cos x = 1$

$
[-\dfrac{1}{2}(\cos 2x - \cos 0)][-\dfrac{1}{2}(\cos 2x - \cos 0)] + [\dfrac{1}{2}(\cos 2x + \cos 0)][\dfrac{1}{2}(\cos 2x + \cos 0)] = 1$

$
[-\dfrac{1}{2}(2\cos^2 x - 2)][-\dfrac{1}{2}(2\cos^2 x - 2)] + [\dfrac{1}{2}(2\cos^2 x)][\dfrac{1}{2}(2\cos^2 x)] = 1$

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

$2\cos^4 x - 2\cos^2 x + 1 = 1$

$\cos^4 x - \cos^2 = 0$

We set u = cos^2 and get $u^2 - u = 0$ which gives us u = 0 and u = 1.

cos^2 x = 0 gives us cos x = +/- 0. x = pi/2 + k(pi).
cos^2 x = 1 gives us cos x = +/- 1. x = 0 + k(pi).

So the solutions are, in order,
0, pi/2, pi, 3pi/2.

I apologize for the mix and match between Latex and regular text.
This is easier if you make use of the Pythagorean Identity...

$\sin^4{x} + \cos^4{x} = 1$

$(\sin^2{x})^2 + (\cos^2{x})^2 = 1$

$(\sin^2{x})^2 + (1 - \sin^2{x})^2 = 1$

$(\sin^2{x})^2 + 1 - 2\sin^2{x} + (\sin^2{x})^2 = 1$

$2(\sin^2{x})^2 - 2\sin^2{x} = 0$

$2\sin^2{x}(\sin^2{x} - 1) = 0$.

Case 1: $\sin^2{x} = 0$

$\sin{x} = 0$

$x = \pi n$.

Case 2: $\sin^2{x} - 1 = 0$

$\sin^2{x} = 1$

$\sin{x} = \pm 1$

$x = \frac{\pi}{2} + \pi n$.

So the solutions are:

$x = \left\{0, \frac{\pi}{2}\right\} + \pi n$

$x = \left\{ 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi \right\}$

(I can't read if you are allowing $2\pi$ to be in the interval... If it's not then get rid of it).

3. Ah, right. Thanks.