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 $\displaystyle \cos^2x$ or $\displaystyle \cos 2x$ ?

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

3. Originally Posted by Moo
Hello,

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

Is it $\displaystyle \sin^2x$ or $\displaystyle \sin 2x$ ?
I was thinking it was $\displaystyle cos2x$ and $\displaystyle sin2x$.... 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
$\displaystyle \cos^2(x)-(1-\cos^2(x))=\cos(x)$

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

Now we just need to solve

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

$\displaystyle \cos(x)-1=0$

Good luck.

9. Ok.

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

~~~~~~~~
Why ?

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

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

The equation to solve is now :

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

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

Try to find the solutions, it has nothing difficult

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

Other method, maybe longer, but I prefer it :

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

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

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

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

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

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. $\displaystyle \cos^2{x}-\sin^2{x}=\cos{x}$
$\displaystyle \cos^2{x}+\sin^2{x}=1$
So,
$\displaystyle \sin^2{x}=1-\cos^2{x}$
Substitute in the original equation:
$\displaystyle \cos^2{x}-(1-\cos^2{x})=\cos{x}$
$\displaystyle \cos^2{x}-1+\cos^2{x}=\cos{x}$
$\displaystyle 2\cos^2{x}-1=\cos{x}$
$\displaystyle 2\cos^2{x}-\cos{x}-1=0$
$\displaystyle (2\cos{x}+1)(\cos{x}-1)=0$
$\displaystyle \cos{x}=-0.5,\;1$
First case:
$\displaystyle \cos{x}=-0.5$
$\displaystyle x=\pm\frac{2\pi}{3}+2n\pi$
$\displaystyle x=\frac{2\pi}{3},\;\frac{4\pi}{3}$
Second case:
$\displaystyle \cos{x}=1$
$\displaystyle x=\frac{\pi}{2}+2n\pi$
$\displaystyle x=\frac{\pi}{2}$

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