Question

Solve for x in the interval [0,2π]:

log₂cot(x) - 2log₄csc(2x) = log₂cos(x)

Answer 1

Answer:

$x=\frac{\pi}{3}$

Step-by-step explanation:

The given logarithmic equation is $\log_2\cot x-2\log_4\csc 2x=\log_2\cos x$.

We regroup to get:

$\log_2\cot x-\log_2\cos x=2\log_4\csc 2x$.

Apply the quotient property of logarithms on the LHS.

$\log_2{\frac{\cot x}{\cos x}=2\log_4\csc 2x$.

Apply the power rule and change of base on the RHS

$\log_2{\frac{1}{\sin x}=\frac{\log_2\csc^2 2x}{\log_24}$.

$\log_2 \csc x=\frac{\log_2\csc^2 2x}{2}$.

$\log_2 \csc x=\frac{\log_2\csc^2 2x}{2}$.

$\log_2 \csc x=\log_2\csc 2x$.

We equate the arguments to get:

$\csc 2x=\csc x$

$\csc 2x-\csc x=0$

Or

$\frac{1}{\sin 2x}-\frac{1}{\sin x}=0$

$\implies \frac{1}{2\sin x\cos x}-\frac{1}{\sin x}=0$

$\frac{1-2\cos x}{2\sin x \cos x}=0$

$\implies 1-2\cos x=0$

$\implies \cos x=0.5$

$\implies x=\frac{\pi}{3},\frac{5\pi}{3}$

But $x=\frac{5\pi}{3}$ is an extraneous solution.

Therefore  $x=\frac{\pi}{3}$ is the only solution in the interval [0,2π]