Question

Precalculus @Hero

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

I think you might know what to do from there

Answer 5

I keep doing these problems linearly.

Answer 6

Like straight algebra

Answer 7

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

Answer 8

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.

Answer 9

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

Answer 10

Thank you again Hero

Answer 11

Yup, it definitely helps.