Solve : cos3x + sin2x – sin4x = 0

Solution :

we have, cos3x + (sin2x – sin4x) = 0

= cos3x – 2sinx.cos3x = 0

\(\implies\)  (cos3x)(1 – 2sinx) = 0

\(\implies\)  cos3x = 0  or  sinx = \(1\over 2\)

\(\implies\)  cos3x = 0 = cos\(\pi\over 2\)  or  sinx = \(1\over 2\) = sin\(\pi\over 6\)

\(\implies\)  3x = 2n\(\pi\) \(\pm\) \(\pi\over 2\)  or  x = m\(\pi\) + \({(-1)}^m\)\(\pi\over 6\)

\(\implies\)  x = \(2n\pi\over 3\) \(\pm\) \(\pi\over 6\)  or  x = m\(\pi\) + \({(-1)}^m\)\(\pi\over 6\); (n, m \(\in\) I)


Similar Questions

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

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

Solve : 6 – 10cosx = 3\(sin^2x\)

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

Leave a Comment

Your email address will not be published.