Question

How do you solve $2{\sin ^2}x = 5\sin x + 3$?

Answer 1

Hint: We will first assume that y = sin x, now we will get a quadratic equation in y. We will now use the method of splitting the middle term and thus get the values of y. After that we will replace y as sin x and thus get the values of x.

Complete step-by-step answer:
We are given that we are required to solve $2{\sin ^2}x = 5\sin x + 3$.
Let us assume that y = sin x.
Replacing this in the given equation, we will obtain the following equation:-
$ \Rightarrow 2{y^2} = 5y + 3$
Taking all the terms from the right hand side to left hand side only to obtain the following expression:-
$ \Rightarrow 2{y^2} - 5y - 3 = 0$
Now, we can write the equation in the above line as following also:-
$ \Rightarrow 2{y^2} + y - 6y - 3 = 0$
Now, taking y common from the first two terms in the left hand side of above expression, we will then obtain the following expression:-
$ \Rightarrow y\left( {2y + 1} \right) - 6y - 3 = 0$
Now, taking - 3 common from the last two terms in the left hand side of the above expression, we will then obtain the following expression:-
$ \Rightarrow y\left( {2y + 1} \right) - 3\left( {2y + 1} \right) = 0$
Now taking the factor (2y + 1) common from the above equation, we will then obtain the following expression:-
$ \Rightarrow \left( {2y + 1} \right)\left( {y - 3} \right) = 0$
Now, we have got two values of y which are $ - \dfrac{1}{2}$ and 3.
Putting back y as sin x as we assumed in the beginning, we will then obtain that the two possible values of sin (x) are $ - \dfrac{1}{2}$ and 3. But since we know that sine of any function always lies between – 1 and 1. Therefore, we can discard the possibility of sin x being 3.
So, we get: $\sin x = - \dfrac{1}{2}$

Therefore, the value of x is $ - \dfrac{\pi }{6}$.

Note:
The students must note that it is very important to discard the value so that we get the actual possible value as here we discarded the possibility that sin x = 3.
The students must also note that in the given solution we used splitting of middle term method but you may also use the fact that the roots of $a{y^2} + by + c = 0$ are given by:-
$ \Rightarrow y = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$