Question

If $1+{\sin }^2\theta =3\sin \theta \cos \theta$ then the solution set in $\left[0,\frac{\pi }{2}\right]$ is

• A. $\{\frac{\pi }{4},{\cos }^{-1}(1/3\left)\right\}$
• B. $\{\frac{\pi }{4},ta{n}^{-1}(1/2\left)\right\}$
• C. $\{\frac{\pi }{3},{\tan }^{-1}(1/3\left)\right\}$
• D. $\{\frac{\pi }{6},si{n}^{-1}(1/3\left)\right\}$

Answer 1

Answer: (B)

$\{\frac{\pi }{4},ta{n}^{-1}(1/2\left)\right\}$

$1+{\sin }^2\theta =3\sin \theta \cos \theta$

Divided by ${\cos }^2\theta$

${\sec }^2\theta +\frac{{\sin }^2\theta }{{\cos }^2\theta }=3\tan \theta$

$1+{\tan }^2\theta +{\tan }^2\theta =3\tan \theta$

$2{\tan }^2\theta -3\tan \theta +1=0$

$2{\tan }^2\theta -2\tan \theta -\tan \theta +1=0$

$2\tan \theta (\tan \theta -1)-(\tan \theta +1)=0$

$(\tan \theta -1)(2\tan \theta -1)=0$

$\tan \theta =1,2\tan \theta =1$

$\theta =\frac{\pi }{4},\theta ={\tan }^{-1}\frac{1}{2}$

$\left\{\frac{\pi }{4},{\tan }^{-1}\left(\frac{1}{2}\right)\right\}$