Logarithms and Trigonometric Functions

Use $\color{#20A900}{\small{\log_{a^2} b=\dfrac2{\log_b a}}}$ and $\color{#20A900}{\small{(\tan x)^{-1}=\cot x}}$ .

$2\log_{\sin x }\cot x - \log_{\cos x} \cot x=0$ $\implies \dfrac{2}{\log_{\cot x} \sin x}-\dfrac1{\log_{\cot x} \cos x}=0$

$\implies \dfrac{2\log_{\cot x}\cos x}{\cancel{\log_{\cot x} \cos x\log_{\cot x}\sin x}}=\dfrac{\log_{\cot x}\sin x}{\cancel{\log_{\cot x} \cos x\log_{\cot x}\sin x}}$ $\implies \cos^2x =\sin x=\sqrt {1-\cos^2 x}$ Squaring and rearranging gives: $\cos^4x+\cos^2x+1=\boxed 2$

$\begin{aligned} \log_{\sin x} \cot^2 x + \log_{\cos x} {\tan x} & = 0 \quad \quad \small \color{#3D99F6}{\text{Changing to a common log base}}\\ \frac {2\log (\cot x)}{\log (\sin x)} + \frac {\log (\tan x)}{\log (\cos x)} & = 0 \\ \frac {2(\log (\cos x) - \log (\sin x))}{\log (\sin x)} + \frac {\log (\sin x) - \log (\cos x)}{\log (\cos x)} & = 0 \\ 2\frac {\log (\cos x)}{\log (\sin x)} - 2 + \frac {\log (\sin x)}{\log (\cos x)} - 1 & = 0 \quad \quad \small \color{#3D99F6}{\text{Multiplying } \frac {\log (\cos x)}{\log (\sin x)} \text{ throughout}} \\ 2 \left(\frac {\log (\cos x)}{\log (\sin x)}\right)^2 - 3\frac {\log (\cos x)}{\log (\sin x)} + 1 & = 0 \\ \left(2\frac {\log (\cos x)}{\log (\sin x)} - 1 \right) \left(\frac {\log (\cos x)}{\log (\sin x)} - 1 \right) & = 0 \\ \implies \frac {\log (\cos x)}{\log (\sin x)} & = \frac 12 \quad \quad \small \color{#3D99F6}{\text{Since } \sin x \ne \cos x} \\ \implies \cos^2 x & = \sin x \end{aligned}$

$\begin{aligned} \implies \cos^4 x + \cos^2 x + 1 & = \sin^2 x + \cos^2 x + 1 \\ & = 1 + 1 = \boxed{2} \end{aligned}$