Question

When $0^o<\theta <{90}^o$, solve the equation ${\cos }^2\theta -3\cos \theta +2={\sin }^2\theta$.

Answer 1

We have,

$\begin{array}{c}{\cos }^2\theta -3\cos \theta +2={\sin }^2\theta \\ \Rightarrow {\cos }^2\theta -3\cos \theta +2=1-{\cos }^2\theta \\ \Rightarrow 2{\cos }^2\theta -3\cos \theta +1=0\\ \Rightarrow 2{\cos }^2\theta -2\cos \theta -\cos \theta +1=0\\ \Rightarrow 2\cos \theta \left(\cos \theta -1\right)-1\left(\cos \theta -1\right)=0\\ \Rightarrow \therefore \cos \theta =\frac{1}{2}or\cos \theta =1\\ \therefore \theta =2h\pi \pm \frac{\pi }{3}or\theta =2k\pi \pm 0,K\in I\end{array}$

Hence, this is the answer.