Question

Determine, for what value of $\theta (0^o\le \theta \le {90}^o),2{\cos }^2\theta -3cos\theta +1=0$

Answer 1

By quadratic formula, $\cos \theta =\frac{3+\sqrt{9-4\left(2\right)}}{4}=\frac{3+1}{4}=\frac{4}{4}$ or $\frac{2}{4}$

$=1$ or $\frac{1}{2}$

$\Rightarrow \cos \theta =1$ or $\cos \theta =\frac{1}{2}$

$\Rightarrow \theta =0^o$ or $\theta ={60}^o$.