Given that \(\sec x + \tan x = \frac{4}{3},\) enter all possible values of \(\sin x\).
+118608 Given
\(secx+tanx=\frac{4}{3}\\ \frac{1}{cosx}+\frac{sinx}{cosx}=\frac{4}{3}\\ 3+3sinx=4cosx\\ (3+3sinx)^2=(4cosx)^2\\ 9+9sin^2x+18sinx=16cos^2x\\ 9+9sin^2x+18sinx=16(1-sin^2x)\\ 9sin^2x+18sinx+9=16-16sin^2x\\ 25sin^2x+18sinx-7=0\\ \text{solve using the quadratic formula and you get}\\ sinx=-1,\frac{7}{25}\\ \text{Using substitution -1 can be ruled out}\\ sinx=\frac{7}{25} \)
