log2(-3sinx)=2log2(cosx)+1 solve for x

Answer 1

log 2 (-3sin x) -log2 (cos^2) x =1

log 2 [ -3 sin x/cos^2 x = 1

-3 sin x / cos^2 x = 2^1 = 2

sin x/cos^2 x = -2/3

well the sin must be negative so x is between pi and 2 pi

well try
sin x/(1 - sin^2 x ) = -2/3
let z = sin x
z/(1-z^2) = -2/3

3 z = -2 + 2 z^2

2 z^2 - 3 z - 2 = 0

(2z+1)(z-2) = 0
z = 2, (not possible, sin >1?)
z = -1/2
ah ha
-30 degrees or - pi/6
or pi + pi/6 = 7 pi/6

Answer 2

Cute problem :)