Question

The number of roots of $(1-\tan \theta )(1+\sin 2\theta )=1+\tan \theta$ for $\theta \in [0,2\pi )]$ is:

• A. $3,3\pi$
• B. $4,2\pi$
• C. $5,\pi$
• D. $(0,2\pi )$

Answer 1

The correct option is C $5,\pi$

$(1-\tan \theta )(1+\sin 2\theta )=(1+\tan \theta )\Rightarrow (1-\frac{\sin \theta }{cos\theta })(1+2\sin \theta \cos \theta )=(1+\frac{\sin \theta }{\cos \theta })\Rightarrow 2{\cos }^2\theta \sin \theta -2{\sin }^2\theta \cos \theta -2\sin \theta =0\Rightarrow 2(1-{\sin }^2\theta )\sin \theta -2{\sin }^2\theta \cos \theta -2\sin \theta =0\Rightarrow {\sin }^3\theta +{\sin }^2\theta \cos \theta =0\Rightarrow {\sin }^2\theta (\sin \theta +\cos \theta )=0\Rightarrow {\sin }^2\theta =0or\sin \theta +\cos \theta =0$

$Now,{\sin }^2\theta =0\Rightarrow \theta =0,\pi ,2\pi for\theta \in [0,2\pi ]and\sin \theta +\cos \theta =0\Rightarrow \tan \theta =-1\Rightarrow \theta =\frac{3\pi }{4},\frac{5\pi }{4}for\theta \in [0,2\pi ]$