Question

If log₂ sin x - log₂ cos x - log₂ ( 1 - tan² x) = -1.
Find the value of x.
How many values x can have?

Answer 1

Step-by-step explanation:

$Given: log_{2} \ sin \ x - log_{2} \ cos \ x - log_{2} \ (1 - tan^2x) = -1$

$\therefore \log_{c}(a) - log_{c} b = log_{c}(\frac{a}{b}) = -1$

$\Longrightarrow log_{2} (\frac{sin \ x}{cos \ x}) - log_{2}(1 - tan^2x) = -1$

$\Longrightarrow log_{2}(\frac{tan(x)}{1-tan^2x}) = -1$

$\Longrightarrow log_{2}(\frac{tan(x)}{1-tan^2x}) = log_{2}(\frac{1}{2})$

$\Longrightarrow \frac{tan(x)}{1-tan^2x} = \frac{1}{2}$

$\Longrightarrow \frac{2tanx}{1-tan^2x} = 1$

$\Longrightarrow tan \ 2x = 1$

$\Longrightarrow tan \ 2x = n\pi + \frac{\pi}{4}$

$\Longrightarrow \boxed{x = \frac{n\pi}{2} + \frac{\pi}{8}}$

Hope it helps!

Answer 2

Notice that the left side of equation has common bases of logs.  So we use the properties of logarithms to write a single log expression on the left side, then equal that to -1.  But before we do so, we will need to rewrite the equation so that it will be easier to apply those properties.

log2sinx - [log2cosx + log2(1 - tan2x)] = -1

Apply the properties of logs in the bracketed terms first.

log2sinx - log2(cosx(1 - tan2x)) = -1

Notice that we started with 3 log terms on the left side.  Now, we only have 2 log terms.  Apply the properties of logs once more to get only one log term.

log2[sinx / (cosx(1 - tan2x))] = -1

The solution to a logarithm is the exponent to a log's base.

2-1 = sinx / [cosx(1 - tan2x)

Then notice that sin(x)/cos(x) is tan(x).

1/2 = tanx / (1 - tan2x)