Find the general solution of (2 sin x - cos x) (1 + cos x) = sin2 x
+
(2n + 1)
We can either expand L.H.S. and proceed or write sin2 x = 1 - cos2 x and proceed. We write
sin2 x = 1 - cos2x, because there is one (1 + cos x) on L.H.S and we can cancel a common factor. We will
proceed in this way because; we don't have to expand L.H.S.
(2 sin x - cos x)(1 + cos x) = 1 - cos2x
= (1 + cos x) (1 - cos x)
(One important point to note here is that when we cancel (1+ cos x) from both the sides, we should equate it to zero. Otherwise we will miss one solution)
⇒ 2 sin x - cos x = 1 - cos x or 1+ cos x = 0
⇒ sin x = 12 or cos x = - 1
⇒ x=nπ + (−1)n π6 or x = (2n + 1) π
Key concepts/6 steps:
(1) sin2 x = 1 - cos2 x
(2) Equating the common factor to zero
(3) sin x = sinα
⇒x=nπ+(−1)nαand
cos x = -1
x = (2n+1) π