A) 0
B) 2
C) 4
D) infinite
Correct Answer: C
Solution :
[b] \[{{\log }_{e}}({{\sec }^{2}}\theta +\cos e{{c}^{2}}\theta )={{\log }_{e}}(4\sin \theta )+lo{{g}_{e}}(16\cos \theta )\]\[\Rightarrow \,\,{{\sec }^{2}}\theta +\cos e{{c}^{2}}\theta =64\sin \theta cos\theta \] \[\Rightarrow \,\,1=64{{\sec }^{3}}\theta {{\cos }^{3}}\theta \] \[\Rightarrow \,\,\,\sin \theta \cos \theta =\frac{1}{4}\] \[\Rightarrow \,\,\,\sin 2\theta =\frac{1}{2}\] \[\Rightarrow \,\,\,2\theta =\frac{\pi }{6},\frac{5\pi }{6},\frac{13\pi }{6},\frac{17\pi }{6}\] \[\Rightarrow \,\,\,\theta =\frac{\pi }{12},\frac{5\pi }{12}\] \[\left( For\,\,\theta =\frac{13\pi }{12},\frac{17\pi }{12};\,\sin \theta ,\cos \theta <0 \right)\]
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