Question

Find the general solution of (2 sin x - cos x) (1 + cos x) = $si{n}^2$ x

• A. $n\pi$ + $(-1{)}^n$ $\frac{\pi }{6}$
• B. (2n + 1)$\pi$
• C. 2n$\pi$
• D. $n\pi$ + $(-1{)}^n$ $\frac{\pi }{3}$

Answer 1

Answer: (A), (B)

(2n + 1)$\pi$

We can either expand L.H.S. and proceed or write $si{n}^2$ x = 1 - $co{s}^2$ x and proceed. We write

$si{n}^2$ x = 1 - $co{s}^2$x, 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 - $co{s}^2$x

= (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)
$\Rightarrow$ 2 sin x - cos x = 1 - cos x or 1+ cos x = 0
$\Rightarrow$ sin x = $\frac{1}{2}$ or cos x = - 1
$\Rightarrow$ $x=n\pi$ + $(-1{)}^n$ $\frac{\pi }{6}$ or x = (2n + 1) $\pi$
Key concepts/6 steps:
(1) $si{n}^2$ x = 1 - $co{s}^2$ x

(2) Equating the common factor to zero

(3) sin x = $sin\alpha$

$\Rightarrow x=n\pi +(-1{)}^n\alpha and$
cos x = -1
x = (2n+1) $\pi$