Question

(4/9)^cosx+2(2/3)^cosx-3=0

Answer 1

$( \frac{4}{9} )^{cosx}+2( \frac{2}{3}) ^{cosx}-3=0 \\ (\frac{2}{3})^{cosx}=t,\ =\ \textgreater \ t\ \textgreater \ 0 \\ t^2+2t-3=0 \\ t_1=1,\ t_2=-3\ =\ \textgreater \ t=1 \\ ( \frac{2}{3}) ^{cosx}=1 \\ cos\ x=0 \\ x= \frac{ \pi }{2} + \pi k,\ k \in Z$