Question

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

• A. 0
• B. n
• C. 2n
• D. None of these

Answer 1

The correct option is A $0$

Given:
${\log }_{|\sin x|}(1+\cos x)=2$

For logarithm to be defined,
$|\sin x|\ne 0\&|\sin x|\ne 1\&1+\cos x\ne 0$
$\Rightarrow x\ne k\pi ,k=0,1,\cdots ,n\cdots \left(i\right)\Rightarrow x\ne (2k+1)\frac{\pi }{2},k=0,\cdots ,n-1\cdots \left(ii\right)$
And
$\Rightarrow x\ne (2k+1)\pi ,k=0,1,\mathrm{2...}\cdots \left(iii\right)$
Now take,
$\Rightarrow {\log }_{|\sin x|}(1+\cos x)=2$
$\Rightarrow 1+\cos x=|\sin x{|}^2\Rightarrow {\sin }^{2}x=1+\cos x$
$\Rightarrow 1-{\cos }^{2}x=1+\cos x\Rightarrow {\cos }^{2}x+\cos x=0\Rightarrow \cos x(1+\cos x)=0\Rightarrow \cos x=0\&\cos x=-1\Rightarrow x=(2k+1)\frac{\pi }{2}\&x=(2k+1)\pi ,k\in Z\cdots \left(A\right)$
$\text{Thus, from}\left(i\right),\left(ii\right),\left(iii\right)\&\left(A\right),$
there is no solution which exists for the
given logarithmic equation.