Question

Solve the following equation

$2{\sin }^2\theta =3\cos \theta$ in the interval $0\le \theta \le 2\pi$.

Answer 1

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

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

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

$2{\cos }^2\theta +3\cos \theta -2=0$

lets take $\cos \theta =x$

so, $2x^2+3x-2=0$

$2x^2+4x-x-2=0$

take 2$x$ common from first two terms and -1 common from last two terms

$2x(x+2)-1(x+2)=0$

$(2x-1)(x+2)=0$

so, $x=\frac{1}{2}$ or $x=-2$

put value of $x=\cos \theta$

$\cos \theta =-2$ is not possible

so, $\cos \theta =\frac{1}{2}$

hence $\theta ={60}^{\circ }$ or $\theta ={300}^{\circ }$