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Question: How do you solve \(2{\sin ^2}x = 5\sin x + 3\)?...

How do you solve 2sin⁡2x=5sin⁡x+32{\sin ^2}x = 5\sin x + 3?

Explanation

Solution

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 2sin⁡2x=5sin⁡x+32{\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:-
⇒2y2=5y+3\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:-
⇒2y2−5y−3=0\Rightarrow 2{y^2} - 5y - 3 = 0
Now, we can write the equation in the above line as following also:-
⇒2y2+y−6y−3=0\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:-
⇒y(2y+1)−6y−3=0\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:-
⇒y(2y+1)−3(2y+1)=0\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:-
⇒(2y+1)(y−3)=0\Rightarrow \left( {2y + 1} \right)\left( {y - 3} \right) = 0
Now, we have got two values of y which are −12 - \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 −12 - \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=−12\sin x = - \dfrac{1}{2}

Therefore, the value of x is −π6 - \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 ay2+by+c=0a{y^2} + by + c = 0 are given by:-
⇒y=−b±b2−4ac2a\Rightarrow y = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}