## Precalculus (6th Edition) Blitzer

If $2{{\cos }^{2}}x-9\cos x-5=0$, then $\underline{2\cos x+1}=0$ or $\underline{\cos x-5}=0$. Of these two equations, the equation that has no solution is $\cos x-5=0$.
To evaluate another value, the middle term splitting is used: \begin{align} & 2{{\cos }^{2}}x-9\cos x-5=0 \\ & 2{{\cos }^{2}}x+\cos x-10\cos x-5=0 \\ & \cos x\left( 2\cos x+1 \right)-5\left( 2\cos x+1 \right)=0 \\ & \left( 2\cos x+1 \right)\left( \cos x-5 \right)=0 \end{align} Therefore, the values will be $2\cos x+1$, another value will be $\cos x-5$ and the equation that has no solution will be $\cos x-5=0$. Thus, the values will be $2\cos x+1,\,\cos x-5\,\text{ and }\,\cos x-5=0$.