Question

How do you solve $\sin \left( {\dfrac{x}{2}} \right) + \cos x - 1 = 0$ over the interval 0 to $2\pi$ ?

Answer 1

Hint: In this question, we need to solve the trigonometric equation over the given intervals. We solve this using the basic identities of trigonometric ratios. We will make use of the identity given by, $\cos 2x = 1 - 2{\sin ^2}x$. From this we find the expression for $\cos x$ and substitute in the given equation. Then we simplify the equation further to solve it and find out the values for the variable x in the given interval and obtain the required solution.

Complete step-by-step answer:
Given an trigonometric expression of the form $\sin \left( {\dfrac{x}{2}} \right) + \cos x - 1 = 0$ …… (1)
We are asked to solve the above expression given in the equation (1) over the interval 0 to $2\pi $.
Firstly, we will find out the expression for cosine in terms of sine. So that it will be easier to simplify and obtain the solution.
We have the identity, $\cos 2x = 1 - 2{\sin ^2}x$
$ \Rightarrow \cos x = 1 - 2{\sin ^2}\left( {\dfrac{x}{2}} \right)$
Substituting this in the equation (1), we get,
$ \Rightarrow \sin \left( {\dfrac{x}{2}} \right) + \left( {1 - 2{{\sin }^2}\left( {\dfrac{x}{2}} \right)} \right) - 1 = 0$
Rearranging the terms, we get,
$ \Rightarrow \sin \left( {\dfrac{x}{2}} \right) + \left( { - 2{{\sin }^2}\left( {\dfrac{x}{2}} \right)} \right) + 1 - 1 = 0$
$ \Rightarrow \sin \left( {\dfrac{x}{2}} \right) + \left( { - 2{{\sin }^2}\left( {\dfrac{x}{2}} \right)} \right) = 0$
Taking out the common factor which is $\sin \left( {\dfrac{x}{2}} \right)$, we get,
$ \Rightarrow \sin \left( {\dfrac{x}{2}} \right)\left( {1 - 2\sin \left( {\dfrac{x}{2}} \right)} \right) = 0$
$ \Rightarrow \sin \left( {\dfrac{x}{2}} \right) = 0$ or $\left( {1 - 2\sin \left( {\dfrac{x}{2}} \right)} \right) = 0$
Given that we need to choose values for x over the intervals 0 to $2\pi $.
If $\sin \left( {\dfrac{x}{2}} \right) = 0$. We know that $\sin 0 = 0$ and $\sin (\pi ) = 0$
So we get, $\sin \left( {\dfrac{x}{2}} \right) = \sin 0$
$ \Rightarrow \dfrac{x}{2} = 0$
$ \Rightarrow x = 0$
Also we have $\sin \left( {\dfrac{x}{2}} \right) = \sin \pi $
$ \Rightarrow \dfrac{x}{2} = \pi $
$ \Rightarrow x = 2\pi $
If $\left( {1 - 2\sin \left( {\dfrac{x}{2}} \right)} \right) = 0$. Then we simplify this equation, we get,
$ \Rightarrow 1 = 2\sin \left( {\dfrac{x}{2}} \right)$
$ \Rightarrow \sin \left( {\dfrac{x}{2}} \right) = \dfrac{1}{2}$
We know that $\sin \dfrac{\pi }{6} = \dfrac{1}{2}$ and also $\sin \dfrac{{5\pi }}{6} = \dfrac{1}{2}$
So we get, $\sin \left( {\dfrac{x}{2}} \right) = \sin \dfrac{\pi }{6}$
$ \Rightarrow \dfrac{x}{2} = \dfrac{\pi }{6}$
Simplifying this, we get,
$ \Rightarrow x = \dfrac{{2\pi }}{6}$
$ \Rightarrow x = \dfrac{\pi }{3}$
Also we have $\sin \left( {\dfrac{x}{2}} \right) = \sin \dfrac{{5\pi }}{6}$
$ \Rightarrow \dfrac{x}{2} = \dfrac{{5\pi }}{6}$
Taking 2 to the other side, we get,
$ \Rightarrow x = \dfrac{{2 \times 5\pi }}{6}$
$ \Rightarrow x = \dfrac{{5\pi }}{3}$
Hence we get the values of x as $x = 0,$ $2\pi ,$ $\dfrac{\pi }{3},$ $\dfrac{{5\pi }}{3}$.
Thus, the solution for the equation $\sin \left( {\dfrac{x}{2}} \right) + \cos x - 1 = 0$ over the interval 0 to $2\pi$ is given by, $x = 0,$ $2\pi ,$ $\dfrac{\pi }{3},$ $\dfrac{{5\pi }}{3}$.

Note:
We can check whether the obtained values of the variable x is correct by substituting back in the given expression. If the equation satisfies the values of x, i.e. if we get L.H.S. is equal to R.H.S. then our value of x is correct. If the equation is not satisfied, then our solution is wrong.
We must know the values of trigonometric ratios for different angles, as it plays an important role in solving such problems and it makes our calculation easier.
Some of the identities of trigonometry are given below.
(1) $\cos 2x = {\cos ^2}x - {\sin ^2}x$
(2) $\cos 2x = 2{\cos ^2}x - 1$
(3) $\cos 2x = 1 - 2{\sin ^2}x$
(4) $\cos 2x = \dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}}$
(5) $\sin 2x = 2\sin x\cos x$