If A is an angle such that (tan A) + (sec A)=3 enter all possible values of cos A separated by commas.
+526 Solution -
\(tan A + sec A = 3 \)
\({sin A\over cos A} + {1\over cos A} = 3 \)
\(sin A + 1 = 3cos A\)
\(sin A = 3cos A - 1\)
sin2A = 9cos2A - 6cos A + 1
1 - cos2A = 9cos2A - 6cos A +1
10cos2A - 6cos A = 0
5cos A - 3 = 0
cos A = 3/5
