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)
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