A) 0
B) 1
C) 2
D) 3
Correct Answer: C
Solution :
\[tan\text{ }x+sec\text{ }x=2\text{ }cos\text{ }x\] \[\frac{\sin x}{\cos x}+\frac{1}{\cos x}=2\cos x\] \[\Rightarrow \] \[\sin x+1=2{{\cos }^{2}}x\] \[\Rightarrow \] \[\sin x+1=2(1-{{\sin }^{2}}x)\] \[\Rightarrow \] \[2{{\sin }^{2}}x+\sin x-1=0\] \[\Rightarrow \] \[(1+\sin x)(2\sin x-1)=0\] \[\Rightarrow \] \[\sin x=-1\]or \[\sin x=\frac{1}{2}\] But the given equation is meaningful only when \[\cos x\ne 0\] ie, \[sin\,x\ne 1\]or \[-1\] Thus, \[\sin x=\frac{1}{2}\] \[\Rightarrow \] \[x=\frac{\pi }{6},\frac{5\pi }{6}\]in \[[0,2\pi ]\] Hence, the given equation has two solutions.
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