Question

Solve $(2+\sqrt{3})\cos \theta =1-\sin \theta$.

• A. $\theta =-2n\pi -\frac{2\pi }{3}$
  
• B. $\theta =2r\pi -\frac{2\pi }{3}$.
• C. $\theta =2r\pi +\frac{2\pi }{3}$.
• D. $\theta =n\pi -\frac{2\pi }{3}$
  

Answer 1

Answer: (A), (B)

The correct options are
A $\theta =-2n\pi -\frac{2\pi }{3}$
B $\theta =2r\pi -\frac{2\pi }{3}$.

$(2+\sqrt{3})\cos \theta =1-\sin \theta$
$\frac{1-\sin \theta }{\cos \theta }=2+\sqrt{3}$
$\frac{{(\cos \frac{\theta }{2}-\sin \frac{\theta }{2})}^2}{{\cos }^2\frac{\theta }{2}-{\sin }^2\frac{\theta }{2}}=2+\sqrt{3}$
or $\frac{1-t}{1+t}=\tan {75}^o,t=\tan \frac{\theta }{2}$
or $\tan (\frac{\pi }{4}-\frac{\theta }{2})=\tan \frac{5\pi }{12}$
$\therefore \frac{\pi }{4}-\frac{\theta }{2}=n\pi +\frac{5\pi }{12}$
or $\theta =-2n\pi -\frac{2\pi }{3}$
or $\theta =2r\pi -\frac{2\pi }{3},r\in I$.