1. Solving trigonometry equation

Solve the equation cos2 x-sin2x=cosx 0≤x<2π

2. Hello,

Originally Posted by kelsey3
Solve the equation cos2 x-sin2x=cosx 0≤x<2π
Is it $\cos^2x$ or $\cos 2x$ ?

Is it $\sin^2x$ or $\sin 2x$ ?

3. Originally Posted by Moo
Hello,

Is it $\cos^2x$ or $\cos 2x$ ?

Is it $\sin^2x$ or $\sin 2x$ ?
I was thinking it was $cos2x$ and $sin2x$.... then i got confused with what i was doing....

4. Its squared

5. Originally Posted by kelsey3
Its squared
For the two ones ?

6. Originally Posted by kelsey3
Its squared
that makes a difference! thank you! i'll see if i can do anything with that.... i bet Moo can figure it out long before me though!

7. ya they're both squared

8. Originally Posted by kelsey3
Solve the equation cos2 x-sin2x=cosx 0≤x<2π
$\cos^2(x)-(1-\cos^2(x))=\cos(x)$

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

Now we just need to solve

$2\cos(x)+1=0$ and

$\cos(x)-1=0$

Good luck.

9. Ok.

$\cos^2x-\sin^2x=\cos(2x)$

~~~~~~~~
Why ?

We know that $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$

$\cos(a+a)=\cos(2a)=\cos^2(a)-\sin^2(a)$
~~~~~~~~

The equation to solve is now :

$\cos(2x)=\cos(x)$

Now, you should know that if $\cos(a)=\cos(b)$, then $a=b+2k \pi$ or $a=-b+2k \pi$

Try to find the solutions, it has nothing difficult

----------------------------------

Other method, maybe longer, but I prefer it :

Remember that $\sin^2 x+\cos^2x=1 \implies \sin^2x=1-\cos^2x$

The equation is now :
$\cos^2x-1+\cos^2x=\cos x$

$2\cos^2x-\cos x-1=0$

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

-> $\cos x-1=0$
or $2 \cos x+1=0$

Edit : too slow

10. aw moo! you were too slow! I bet it's my fault! hehe....

Wow, i never would have gotten that.... But I get it now! Thank you Kelsey3 for askin'!

11. $
\cos^2{x}-\sin^2{x}=\cos{x}
$

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

So,
$
\sin^2{x}=1-\cos^2{x}
$

Substitute in the original equation:
$
\cos^2{x}-(1-\cos^2{x})=\cos{x}
$

$
\cos^2{x}-1+\cos^2{x}=\cos{x}
$

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

$
2\cos^2{x}-\cos{x}-1=0
$

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

$
\cos{x}=-0.5,\;1
$

First case:
$
\cos{x}=-0.5
$

$
x=\pm\frac{2\pi}{3}+2n\pi
$

$
x=\frac{2\pi}{3},\;\frac{4\pi}{3}
$

Second case:
$
\cos{x}=1
$

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

$
x=\frac{\pi}{2}
$

Final solution:
$
x = \frac{2\pi}{3}, \frac{4\pi}{3}, \frac {\pi}{2}
$