A) \[0\]
B) \[1\]
C) \[2\]
D) \[3\]
Correct Answer: C
Solution :
Given equation is \[\tan x+\sec x=2\cos x\] \[\Rightarrow \] \[\sin x+1=2{{\cos }^{2}}x\] \[\Rightarrow \] \[2(1-{{\sin }^{2}}x)=\sin x+1\] \[\Rightarrow \] \[2{{\sin }^{2}}x+\sin x-1=0\] \[\Rightarrow \] \[(2\sin x-1)(\sin x+1)=0\] \[\Rightarrow \] \[\sin x=\frac{1}{2}\] or \[\sin x=-1\] \[\Rightarrow \] \[x=\frac{\pi }{6},\,\,\frac{5\pi }{6}\] or \[x=\frac{3\pi }{2}\] \[\Rightarrow \]\[x=\frac{\pi }{6},\,\,\frac{5\pi }{6}\] \[\left( \because \,\,x=\frac{3\pi }{2},\,\,\tan x\,\,\text{is}\,\,\text{not}\,\,\text{defined} \right)\] Hence, there are two solutions exist.
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