Question

Solve: $3-3\cos \theta =2{\sin }^2\theta$.

Answer 1

$3-3\cos \theta =2{\sin }^2\theta$

$3(1-\cos \theta )=2(1-{\cos }^2\theta )$

$\Rightarrow 3(1-\cos \theta )=2(1-\cos \theta )(1+\cos \theta )$

$\Rightarrow 3(1-\cos \theta )-2(1-\cos \theta )(1+\cos \theta )=0$

$\Rightarrow (1-\cos \theta )[3-2(1+\cos \theta )]=0$

$\Rightarrow (1-\cos \theta )[3-2+2\cos \theta ]=0$

$\Rightarrow (1-\cos \theta )[1+2\cos \theta ]=0$

$\Rightarrow 1-\cos \theta =0$, $1+2\cos \theta =0$

$\Rightarrow \cos \theta =1,\cos \theta =\frac{-1}{2}$

$\Rightarrow \theta =2n\pi +0=2n\pi ,\theta =2n\pi +(\pi -\frac{\pi }{3})$

$\therefore \theta =2n\pi ,2n\pi +\frac{2\pi }{3}$

$\therefore \theta =0,\frac{2\pi }{3}$