Question

If $\theta \in (0,2\pi )$ and $2\cos \theta =\sqrt{3}\cos {10}^{\circ }-\sin {10}^{\circ }$, then $\theta$ can be

• A. ${20}^{\circ }$
• B. ${340}^{\circ }$
• C. ${320}^{\circ }$
• D. ${40}^{\circ }$

Answer 1

Answer: (C), (D)

The correct options are
C ${320}^{\circ }$
D ${40}^{\circ }$

$2\cos \theta =\sqrt{3}\cos {10}^{\circ }-\sin {10}^{\circ }\Rightarrow \cos \theta =\frac{\sqrt{3}}{2}\cos {10}^{\circ }-\frac{1}{2}\sin {10}^{\circ }\Rightarrow \cos \theta =\cos {30}^{\circ }\cos {10}^{\circ }-\sin {30}^{\circ }\sin {10}^{\circ }\Rightarrow \cos \theta =\cos ({30}^{\circ }+{10}^{\circ })\Rightarrow \cos \theta =\cos \left({40}^{\circ }\right)=\cos \left({320}^{\circ }\right)\therefore \theta ={40}^{\circ }={320}^{\circ }$