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\(\pi\) \(\pm\) \(cos^{-1}{1\over 3}\)


Similar Questions

If \({1\over 6}sin\theta\), \(cos\theta\) and \(tan\theta\) are in G.P. then the general solution for \(\theta\) is

Solve : cos3x + sin2x – sin4x = 0

Find the number of solutions of tanx + secx = 2cosx in [0, \(2\pi\)].

Find general solution of (2sinx – cosx)(1 + cosx) = \(sin^2x\)

Leave a Comment

Your email address will not be published.