Question

How do you find all solutions of $\sin \left( {\dfrac{x}{2}} \right) + \cos x - 1 = 0$ in the interval $[0,2\pi )$?

Answer 1

We will first use the formula that $\cos 2\theta = 1 - 2{\sin ^2}\theta$ and then we will get a quadratic in $\sin \left( {\dfrac{x}{2}} \right)$ which can be solved by taking common since constant is zero in this.

Complete step by step solution:
We are given that we are required to find all the solutions of $\sin \left( {\dfrac{x}{2}} \right) + \cos x - 1 = 0$ in the interval $[0,2\pi )$.
Since, we know that we have a formula given by the following expression:-
$\Rightarrow \cos 2\theta = 1 - 2{\sin ^2}\theta$
Replacing $\theta$ by $\dfrac{x}{2}$ in the above mentioned formula, the following equation is obtained:$\cos x = 1 - 2{\sin ^2}\left( {\dfrac{x}{2}} \right)$
Putting this in the given expression in the question, the following equation is obtained:$\sin \left( {\dfrac{x}{2}} \right) + 1 - 2{\sin ^2}\left( {\dfrac{x}{2}} \right) - 1 = 0$
We can combine 1 and – 1 in the above expression and after doing that, the following equation is obtained:
$\sin \left( {\dfrac{x}{2}} \right) + 2{\sin ^2}\left( {\dfrac{x}{2}} \right) = 0$
Taking $\sin \left( {\dfrac{x}{2}} \right)$ common from both the terms in the above mentioned expression, the following equation is obtained:
$\Rightarrow \sin \left( {\dfrac{x}{2}} \right)\left( {1 + 2\sin \left( {\dfrac{x}{2}} \right)} \right) = 0$
Therefore, the following are the possibilities obtained:-
Either $\sin \left( {\dfrac{x}{2}} \right) = 0$ or $1 + 2\sin \left( {\dfrac{x}{2}} \right) = 0$
This implies that either $\dfrac{x}{2} = 0,n\pi$ or $\sin \left( {\dfrac{x}{2}} \right) = - \dfrac{1}{2}$.
This implies that either $x = 0,2n\pi$ or $\dfrac{x}{2} = 2n\pi - \dfrac{\pi }{6}$.
This implies that either $x = 0,2n\pi$ or $x = 4n\pi - \dfrac{\pi }{3}$.
Considering the interval, the final answer is $0,\dfrac{{11\pi }}{3}$.

Note: The students must note that we have used the fact that: If a.b = 0, then either a = 0 or b = 0.
Thus is true for all the real numbers a and b.
The students must also note that, they could have used the quadratics formula to solve the following equation:-
$\Rightarrow \sin \left( {\dfrac{x}{2}} \right) + 2{\sin ^2}\left( {\dfrac{x}{2}} \right) = 0$
If we have the general quadratic equation as $a{x^2} + bx + c = 0$, then its roots are given by $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Comparing it, we have a quadratic in $\sin \left( {\dfrac{x}{2}} \right)$, where a = 2, b = 1 and c = 0.
Therefore, the roots of the equation are:-
$\Rightarrow \sin \left( {\dfrac{x}{2}} \right) = \dfrac{{ - 1 \pm \sqrt 1 }}{4}$
Simplifying the calculations above, we have the following equation:-
$\Rightarrow \sin \left( {\dfrac{x}{2}} \right) = 0, - \dfrac{1}{2}$
This implies that either $\dfrac{x}{2} = 0,n\pi$ or $\sin \left( {\dfrac{x}{2}} \right) = - \dfrac{1}{2}$.
Thus, we have either $x = 0,2n\pi$ or $x = 4n\pi - \dfrac{\pi }{3}$.
Considering the interval, the final answer is $0,\dfrac{{11\pi }}{3}$.