Solve : 6 - 10cosx = 3\(sin^2x\)

Solution :

we have, 6 – 10cosx = 3\(sin^2x\)

\(\therefore\) 6 – 10cosx = 3 – 3\(cos^2x\)

\(\implies\) 3\(cos^2x\) – 10cosx + 3 = 0

\(\implies\) (3cosx-1)(cosx-3) = 0 \(\implies\) cosx = \(1\over 3\) or cosx = 3

Since cosx = 3 is not possible as -1 \(\le\) cosx \(\le\) 1

\(\therefore\) cosx = \(1\over 3\) = cos(\(cos^{-1}{1\over 3}\)) \(\implies\) x = 2n\(\pi\) \(\pm\) \(cos^{-1}{1\over 3}\)