Question

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

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

Answer 1

The correct option is A $0$

$n\in Z$
For ${\log }_a\left(x\right);x,a>0$ but $a\ne 1$
Using this, we can say that $\left|\sin \right(x\left)\right|>0and\left|\sin \right(x\left)\right|\ne 1$, which gives us $x=[0,n\pi ]-\left(\frac{n\pi }{2}\right)$......... {1}
${\log }_a\left(x\right)=y$ gives $a^y=x$.............. (logarithmic property)
Using this in expression ${\log }_{|\sin \left(x\right)|}(1+\cos (x\left)\right)=2$, we get
$|\sin \left(x\right){|}^2=1+\cos \left(x\right)$
$\to$ $1-{\cos }^2\left(x\right)=1+\cos \left(x\right)$
$\to$ $\cos \left(x\right)\ast (1+\cos (x\left)\right)=0$ gives either $\cos \left(x\right)=0$ or $\cos \left(x\right)=-1$, but $\cos \left(x\right)\ne -1$ because it violates logarithmic property
and for $\cos \left(x\right)=0$, $x$ has to be $\frac{n\pi }{2}$ which is not possible...... from {1}
$\therefore$ there is no values of $x$ that satisfies the expression.
Hence option A.