Question

log(2)(2sinx)+log(2)(cosx)=-1

Answer 1

Assuming the equation is

$\log_22\sin x+\log_2\cos x=-1$

and that you're solving for real $x$ only. Recall that $\log_2y$ is defined for $y>0$, which means we have to have $2\sin x>0$ and $\cos x>0$, both of which happen so long as $0<x<\dfrac\pi2$.

First we can contract the logarithms into one, then apply a double angle identity:

$\log_22\sin x+\log_2\cos x=\log_22\sin x\cos x=\log_2\sin2x$

Then we treat both sides as equal powers of 2:

$2^{\log_2\sin2x}=2^{-1}\implies\sin2x=\dfrac12$

which occurs for $2x=\dfrac\pi6+2n\pi$ or $2x=\dfrac{5\pi}6+2n\pi$ for any integer $n$. So $x=\dfrac\pi{12}+n\pi$ or $x=\dfrac{5\pi}{12}+n\pi$.

Remember that we need to have $0<x<\dfrac\pi2$, so we have to omit some solutions. Both the "fundamental" solutions $\dfrac\pi{12}$ and $\dfrac{5\pi}{12}$ fall in the desired interval, but if we add or subtract an odd multiple of $\pi$, we fall outside of it. For example, if $n=-1$, then

$2\sin\left(\dfrac\pi{12}-\pi\right)=2\sin\left(-\dfrac{11\pi}{12}\right)=-2\sin\dfrac{11\pi}{12}<0$

which would make $\log_22\sin x$ undefined.

So the overall solution set would be

$x=\dfrac\pi{12}+2k\pi$

or

$x=\dfrac{5\pi}{12}+2k\pi$

for integers $k$.