# Math Help - trig

1. ## trig

4cos²x + cosx = 0

and

2sin²x - sinx - 1 = 0

thanks

2. Originally Posted by gracey
4cos²x + cosx = 0

and

2sin²x - sinx - 1 = 0

thanks
Think of these as "polynomials".

For $4cos^2x+cosx=0$, let $u=cosx$.
From that, you will get $4u^2+u=0$. Factor and solve that for u, then plug $cosx$ back in for u.

Take the same approach for the second problem.

3. First convert them to quadratic functions.

Let $a=\cos x$

$4\cos^2x + cosx = 0$

$4a^2 + a = 0$

$a(4a+1) = 0$

1. $a=0$
2. $4a+1=0$, $a=-\frac{1}{4}$

$a=0$
$\cos x = 0$
$\boxed{~x = \frac{\pi}{2} + k\pi~}$

$a=-\frac{1}{4}$
$\cos x = -\frac{1}{4}$
$\boxed{~x=-\arccos \left(-\frac{1}{4}\right)+2 k \pi}$
$\boxed{~x=\arccos \left(-\frac{1}{4}\right)+2 k \pi}$