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

Hint: 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 \[0,\dfrac{{11\pi }}{3}\].