Question

Find the values of x for which $\sqrt{1-\cos x}=\sin x$, where $n\in I$
A. $\left( 2n+1 \right)\dfrac{\pi }{2}$
B. $2n\pi +\dfrac{\pi }{4}\text{ or }n\pi $
C. $n\pi \text{ or }\left( 2n+1 \right)\dfrac{\pi }{4}$
D. none of the above

Answer 1

Hint: To solve this question, we have to square the equation $\sqrt{1-\cos x}=\sin x$ on both sides and use the relation ${{\sin }^{2}}x=1-{{\cos }^{2}}x$ and simplifying the equation, we get
${{\cos }^{2}}x-\cos x=0$. From this equation, we can write $\cos x=0 $ or $\cos x-1=0 $. From these equations, we can write the general solutions as $x=\left( 2n+1 \right)\dfrac{\pi }{2}\text{ or }2n\pi $.

Complete step-by-step solution:
In the question, it is given that $\sqrt{1-\cos x}=\sin x$, and we are asked to find the values of x which satisfy the equation.
To solve these types of questions, we have to remove the square-roots from the equation so that we can simplify them. Squaring on both sides, we get
\[\begin{align}
  & {{\left( \sqrt{1-\cos x} \right)}^{2}}={{\left( \sin x \right)}^{2}} \\
 & 1-\cos x={{\sin }^{2}}x\to \left( 1 \right) \\
\end{align}\]
To simplify further, we have to recall an equation from trigonometry which is
${{\sin }^{2}}x+{{\cos }^{2}}x=1$
From this equation, we can write that
${{\sin }^{2}}x=1-{{\cos }^{2}}x$
Using this in equation-1, we get
\[\begin{align}
  & 1-\cos x={{\sin }^{2}}x=1-{{\cos }^{2}}x \\
 & 1-\cos x=1-{{\cos }^{2}}x \\
\end{align}\]
Cancelling 1 on both sides, we get
$-\cos x=-{{\cos }^{2}}x$
Multiplying by -1 and getting the two terms to one side, we get
${{\cos }^{2}}x-\cos x=0$
Taking one $\cos x$ term outside, we get
$\cos x\left( \cos x-1 \right)=0$
When $ab=0$, we can write that either $a=0$or $b=0$
Here $a=\cos x$ and $b=\cos x-1$
Let us consider case-1 $a=\cos x=0$
For $\cos x=0$ we get the values of x as $x=\dfrac{\pi }{2},\dfrac{3\pi }{2},\dfrac{5\pi }{2},......$
The values of x are odd multiples of $\dfrac{\pi }{2}$
The general solution of $\cos x=0$ can be written as $x=\left( 2n+1 \right)\dfrac{\pi }{2}$ where $n\in I$
Let us consider the case of $b=\cos x-1=0$ which means that $\cos x=1$
We can find the values of x for which $\cos x=1$ , they are given by $x=0,2\pi ,4\pi ....$
The solutions of x are even multiples of $\pi $ , which we can write as
$x=2n\pi $ where $n\in I$
$\therefore $ The values of x for which $\sqrt{1-\cos x}=\sin x$ are given by $x=\left( 2n+1 \right)\dfrac{\pi }{2}\text{ or }2n\pi $. The answer is option-A.

Note: Students might go wrong while choosing the answer as there is no option which completely includes the range of values of x. The option A is the partial solution which contains some values of the solution set $x=\left( 2n+1 \right)\dfrac{\pi }{2}\text{ or }2n\pi $. So, in this case, the answer is option-A.