Precalculus @Hero

3 months ago

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

3 months ago

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

I keep doing these problems linearly.

3 months ago

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

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

3 months ago