MHB Solve $2\cos^2(x)=1$: Find $\theta$

[karush]

$\tiny{\textsf{9c Miliani HS hw}}$
Find x
$2\cos^2(x)=1$
$\cos^2(x)=\dfrac{1}{2}\implies
\cos(x)=\dfrac{1}{\sqrt{2}}\implies
x=\dfrac{\pi}{4}$ or $x=\dfrac{7\pi}{4}$

well its the simple ones where we stumple
not sure if it necessary to give all quadrants of possible answers
its probably better to us $\theta$ rather than x
mahalo

 

[DaalChawal]

You forgot $-1 \over root2$ as the other solution.
There are two methods one $cos^2x = cos^2\alpha$ , $x= n \pi + \alpha$ and $x = n \pi - \alpha$
and second method $cosx = cos\alpha , x = 2n \pi + \alpha , x = 2n \pi - \alpha$
take any alpha which satisfies the value and you will get whole possible answers.

 

[karush]

well that's pretty valuable to know
i was thinking of just one revolution

 

[skeeter]

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

$\cos(2x) = 0$

$2x = (2k+1) \cdot \dfrac{\pi}{2} \implies x = (2k+1) \cdot \dfrac{\pi}{4} \, , \, x \in \mathbb{Z}$

 

[karush]

ok not sure where 2x= came from but see that it works

 

[skeeter]

ok not sure where 2x= came from but see that it works

double angle identity for cosine derivation …

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

also, note …

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

 

[DaalChawal]

He might also don't know about $cos(A+B) = cosA cosB - sinA sinB$ and all other formulas

 

[karush]

yes i know that one its very common