Shadow:

Precalculus @Hero

3 months ago
Shadow:

\[1 - \sin^2 x = 0.5 \cos x\] \[\cos^2 x = 0.5 \cos x\] \[(\cos x) (\cos x) = 0.5 \cos x\] \[\cos x = \frac{ 1 }{ 2 }\] \[x = \frac{ \pi }{ 3 }\]

3 months ago
Shadow:

These cos^2 x always hide a solution, so I know I'm missing something here.

3 months ago
Hero:

\(\cos^2x - \dfrac{1}{2}\cos(x) = 0\) \(\cos(x)\left(\cos(x) - \dfrac{1}{2}\right) = 0\)

3 months ago
Hero:

I think you might know what to do from there

3 months ago
Shadow:

I keep doing these problems linearly.

3 months ago
Shadow:

Like straight algebra

3 months ago
Hero:

Yeah, it's best to think of these like quadratic equations. Move everything to the left and set equal to zero.

3 months ago
Hero:

Technically, if you let \(\cos(x) = x\) then you would have: \(x^2 = 0.5x\) If you thought of it that way, you would know to automatically move everything to the left side: \(x^2 - 0.5x = 0\) \(x(x - 0.5) = 0\) Becomes more obvious then.

3 months ago
Shadow:

That's true. I think I'll start converting my trig functions to x since they look easier that way.

3 months ago
Shadow:

Thank you again Hero

3 months ago
Hero:

Yup, it definitely helps.

3 months ago