Mathematics - Trigonometric Equation Question with Solution | TestHub

${\log }_{|\sin x|}(1+\cos x)=2\Rightarrow 1+\cos x=|\sin x{|}^2$ $\Rightarrow 1+\cos \mathrm{x}=1-{\cos }^2\mathrm{x}\Rightarrow \cos \mathrm{x}(1+\cos \mathrm{x})=0$ $\mathrm{But}(1+\cos \mathrm{x})\ne 0\Rightarrow \cos \mathrm{x}=0,\Rightarrow \sin \mathrm{x}=1$.

But $\sin \mathrm{x}=1$ is not possible because the base of $\log$ can not be 1 .

Hence no solution.