Find the angle satisfying the equation $4\log_{16}{(\cos{2x})}$ $+$ $2\log_{4}{(\sin{x})}$ $+$ $\log_{2}{(\cos{x})}$ $+3$ $\,=\,$ $0$

It is a mathematical equation in which logarithmic and trigonometric functions are involved. The variable $x$ is also used to represent angle in this case. It can be evaluated by solving this logarithmic trigonometric equation.

Write base of log terms in exponential form

The bases of first, second and third logarithmic terms are $16$, $4$ and $2$ respectively. It is possible to express the bases of first and second terms in terms of base of third log term. This technique helps us to combine all three terms in next few steps. So, write the bases of first two terms in exponential notation.

$\implies$ $4\log_{2^4}{(\cos{2x})}$ $+$ $2\log_{2^2}{(\sin{x})}$ $+$ $\log_{2}{(\cos{x})}$ $+3$ $\,=\,$ $0$

Simplify expression by log base power rule

The three log terms can be made as like terms by applying base power rule of logarithms.

$\implies$ $4 \times \dfrac{1}{4} \times \log_{2}{(\cos{2x})}$ $+$ $2 \times \dfrac{1}{2} \times \log_{2}{(\sin{x})}$ $+$ $\log_{2}{(\cos{x})}$ $+3$ $\,=\,$ $0$

Now, simplify all three terms in this logarithmic trigonometric equation.

$\implies$ $\dfrac{4}{4}\log_{2}{(\cos{2x})}$ $+$ $\dfrac{2}{2}\log_{2}{(\sin{x})}$ $+$ $\log_{2}{(\cos{x})}$ $+3$ $\,=\,$ $0$

$\implies$ $\require{cancel} \dfrac{\cancel{4}}{\cancel{4}}\log_{2}{(\cos{2x})}$ $+$ $\require{cancel} \dfrac{\cancel{2}}{\cancel{2}}\log_{2}{(\sin{x})}$ $+$ $\log_{2}{(\cos{x})}$ $+3$ $\,=\,$ $0$

$\implies$ $\log_{2}{(\cos{2x})}$ $+$ $\log_{2}{(\sin{x})}$ $+$ $\log_{2}{(\cos{x})}$ $+$ $3$ $\,=\,$ $0$

Combine the logarithmic terms

The like logarithmic terms can be combined by using product rule of logarithms.

$\implies$ $\log_{2}{(\cos{2x} \times \sin{x} \times \cos{x})}$ $\,=\,$ $-3$

Transform log term into exponential form

The logarithmic equation can be expressed in exponential notation by the mutual relation between them.

$\implies$ $\cos{2x} \times \sin{x}\cos{x}$ $\,=\,$ $2^{-3}$

Now, continue the simplification procedure to simplify the mathematical equation.

$\implies$ $\cos{2x} \times \sin{x}\cos{x}$ $\,=\,$ $\dfrac{1}{2^3}$

$\implies$ $\cos{2x} \times \sin{x}\cos{x}$ $\,=\,$ $\dfrac{1}{8}$

Convert part of the product as sin of double angle

The part of the product represents expansion of sin double angle identity but it requires the number $2$ as its multiple. It can be brought from right hand side quantity.

$\implies$ $\cos{2x} \times \sin{x}\cos{x}$ $\,=\,$ $\dfrac{1}{2 \times 4}$

$\implies$ $2 \times \cos{2x} \times \sin{x}\cos{x}$ $\,=\,$ $\dfrac{1}{4}$

$\implies$ $\cos{2x} \times 2\sin{x}\cos{x}$ $\,=\,$ $\dfrac{1}{4}$

Now, it can be written as $\sin{2x}$ as per sin double angle formula.

$\implies$ $\cos{2x} \times \sin{2x}$ $\,=\,$ $\dfrac{1}{4}$

$\implies$ $\sin{2x} \times \cos{2x}$ $\,=\,$ $\dfrac{1}{4}$

$\implies$ $\sin{2x}\cos{2x}$ $\,=\,$ $\dfrac{1}{4}$

Surprisingly, the product also represents part of the sin double angle identity. So, bring the number $2$ to left hand side from the right hand side quantity for writing it as sin of double angle.

$\implies$ $\sin{2x}\cos{2x}$ $\,=\,$ $\dfrac{1}{2 \times 2}$

$\implies$ $2 \times \sin{2x}\cos{2x}$ $\,=\,$ $\dfrac{1}{2}$

$\implies$ $2\sin{2x}\cos{2x}$ $\,=\,$ $\dfrac{1}{2}$

$\implies$ $\sin{(2 \times 2x)}$ $\,=\,$ $\dfrac{1}{2}$

Now, write the product as the sin of double angle form.

$\implies$ $\sin{4x}$ $\,=\,$ $\dfrac{1}{2}$