1. ## Trig equations

I got stuck in these 2 equations :

$tan^{2}x=\frac{1-cosx}{1-sin\left | x \right |}$
$\left | log_{\frac{1}{3}}(1+sin2x) \right |+\left | log_{\frac{1}{3}}(1-sin2x)\right |=1$

2. Originally Posted by ldx2
I got stuck in these 2 equations :

$tan^{2}x=\frac{1-cosx}{1-sin\left | x \right |}$
$\left | log_{\frac{1}{3}}(1+sin2x) \right |+\left | log_{\frac{1}{3}}(1-sin2x)\right |=1$
second one: you can combine those two expressions into a single log.

$| \log_{1/3}\left[(1+sin[2x])(1-sin[2x])\right) |$

Which is the difference of two squares

$= \log_{1/3} \left(1-sin^2(2x)\right) = \log_{1/3} cos^2(2x)$

This is equal to 1

\log_{1/3} cos^2(2x) = \pm 1

$cos^2(2x) = \frac{1}{3}$ or $cos^2(2x) = 3$

The second solution is not a solution because cos^2(2x) cannot be 3