I need to solve $\displaystyle 4\sin\theta + 5\cos\theta =6$

$\displaystyle 4(1-\cos^2\theta)+5\cos\theta -6=0$

$\displaystyle -4\cos^2\theta +5\cos\theta -2=0$

$\displaystyle \cos\theta = \frac{-5\pm \sqrt{25-4(-2)(-4)}}{8}$

This gives me complex roots. In my book they solve the equation as follows:

$\displaystyle 4\sin\theta=6-5\cos\theta$

$\displaystyle 16\sin^2\theta =36-60\cos\theta +25\cos^2\theta $

$\displaystyle 16-16\cos^2\theta = 36-60\cos\theta +25\cos^2\theta $

$\displaystyle 0=41\cos^2\theta -60\cos\theta +20$

$\displaystyle \cos\theta = \frac{60 \pm\sqrt{3,600-4(41)(20)}}{2(41)}$

=$\displaystyle 0.9499 or 0.5136$

How was my approach to the problem wrong?