Question

Solve $1+\cos x={\sin }^{2}x$

• A. $(2n+1)\frac{\pi }{2}\forall n\in N$
• B. $2n\pi \forall n\in N$
• C. $(2n+1)\pi \forall n\in N$
• D. $n\pi \forall n\in N$

Answer 1

Answer: (A), (C)

The correct options are
A $(2n+1)\frac{\pi }{2}\forall n\in N$
C

$(2n+1)\pi \forall n\in N$

Here, we will write ${\sin }^{2}x=1-{\cos }^{2}x$, because there is a $1+\cos x$ on L.H.S. and it will be a common factor.

$\Rightarrow 1+\cos x=1-co{s}^{2}x$

$\Rightarrow 1+\cos x=(1-\cos x)(1+\cos x)$

$\Rightarrow 1+\cos x=0\text{or}1-\cos x=1$

$\Rightarrow \cos x=-1\text{or}\cos x=0$
$\Rightarrow x=\pi$ or $x=\frac{\pi }{2}$

We know that the general solution forumla for the equation
$\cos x=\cos \alpha$
is $x=2n\pi \pm \alpha$
$\Rightarrow x=(2n\pm 1)\pi$
or $x=(2n\pm 1)\frac{\pi }{2}$