Question

The number of values of $x\in [0,n\pi ],n\in I$ that satisfy ${\log }_{\left|\sin x\right|}(1+\cos x)=2$ is

• A. $0$
• B. $n$
• C. $2n$
• D. none of these

Answer 1

The correct option is A $0$

The equation is meaningful if $\left|\sin x\right|\ne 0,1$ and $1+\cos x\ne 0$
so $x\ne k\pi$ where, $k=0,1,.....n$
and $x\ne (2k+1)\frac{\pi }{2},k=0,1,.....n-1$.
Now ${\log }_{\left|\sin x\right|}(1+\cos x)=2$
$\iff 1+\cos x={\left|\sin x\right|}^2={\sin }^{2}x=1-{\cos }^{2}x$
$\Rightarrow (1+\cos x)\left(\cos x\right)=0$
$\Rightarrow \cos x=0or\cos x=-1$
$\Rightarrow \cos x=0\Rightarrow x=(2k+1)\pi /2$
So there is no $x$ which satisfy the given equation
Ans: A