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<\frac\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=\frac12`

which occurs for
`2x=\frac\pi6+2n\pi` or
`2x=\frac{5\pi}6+2n\pi` for any integer
`n`. So
`x=\frac\pi{12}+n\pi` or
`x=(5\pi)/(12)+n\pi`.

Remember that we need to have
`0<x<\frac\pi2`, so we have to omit some solutions. Both the "fundamental" solutions
`\frac\pi{12}` and
`(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(\frac\pi{12}-\pi\right)=2\sin\left(-(11\pi)/(12)\right)=-2\sin(11\pi)/(12)<0`

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

So the overall solution set would be

`x=\frac\pi{12}+2k\pi`

or

`x=(5\pi)/(12)+2k\pi`

for integers
`k`.