Question

The set of values of $a$ for which the equation $\sin x(\sin x + \cos x) = a$ has real solution is
A. \left\\{ {1 - \sqrt 2 ,1 + \sqrt 2 } \right\\}
B. \left\\{ {2 - \sqrt 3 ,2 + \sqrt 3 } \right\\}
C. \left\\{ {0,2 + \sqrt 3 } \right\\}
D. \left\\{ {\dfrac{{1 - \sqrt 2 }}{2},\dfrac{{1 + \sqrt 2 }}{2}} \right\\}

Answer 1

In this question, we are given an equation in which, there is an unknown constant $a$ , for which we have to find the real solutions, which will satisfy the equation $\sin x(\sin x + \cos x) = a$. First open the brackets using distributive property over addition and convert each term into non- exponential function and then, simplify the resultant.

Formulae to be used:
$- \sqrt 2 < \sin x - \cos x < \sqrt 2$
$\cos 2x = 1 - 2{\sin ^2}x$
$\sin 2x = 2\sin x\cos x$

Complete step-by-step answer:
We are given an equation $\sin x(\sin x + \cos x) = a$ .
To find values of $a$ for which the given equation has real solutions.
Given equation is $\sin x(\sin x + \cos x) = a$ , first we will multiply $\sin x$ with each term using distributive property over addition, we get, ${\sin ^2}x + \sin x\cos x = a$ .
Now, we will convert it into non- exponential function, so, using the identity, $\cos 2x = 1 - 2{\sin ^2}x$ , we get, ${\sin ^2}x = \dfrac{{1 - \cos 2x}}{2}$ , so, replace ${\sin ^2}x$ by this value, the equation becomes, $\dfrac{{1 - \cos 2x}}{2} + \sin x\cos x = a$ .
Now, in $\sin x\cos x$ , multiply and divide by $2$ , we get, $\dfrac{{1 - \cos 2x}}{2} + \dfrac{{2\sin x\cos x}}{2} = a$ .
Now, we have the identity, $\sin 2x = 2\sin x\cos x$ , so replacing $2\sin x\cos x$ by $\sin 2x$ , the equation becomes, $\dfrac{{1 - \cos 2x}}{2} + \dfrac{{\sin 2x}}{2} = a$ , which can be written as, $1 - \cos 2x + \sin 2x = 2a$ i.e., $\sin 2x - \cos 2x = 2a - 1$ .
We have that, $- \sqrt 2 < \sin x - \cos x < \sqrt 2$ , hence, $- \sqrt 2 < \sin 2x - \cos 2x < \sqrt 2$ , so, $- \sqrt 2 < 2a - 1 < \sqrt 2$ .
Now, adding $1$ on each side, we get, $- \sqrt 2 + 1 < 2a - 1 + 1 < \sqrt 2 + 1$ , which can be written as, $1 - \sqrt 2 < 2a < 1 + \sqrt 2$ . Now, finally, dividing each side by $2$ , we get, $\dfrac{{1 - \sqrt 2 }}{2} < \dfrac{{2a}}{2} < \dfrac{{1 + \sqrt 2 }}{2}$ i.e., $\dfrac{{1 - \sqrt 2 }}{2} < a < \dfrac{{1 + \sqrt 2 }}{2}$ .
Hence, interval for values of $a$ for which the equation $\sin x(\sin x + \cos x) = a$ has real solution is \left\\{ {\dfrac{{1 - \sqrt 2 }}{2},\dfrac{{1 + \sqrt 2 }}{2}} \right\\} .

So, the correct answer is “Option D”.

Note: The property which we have used in this question, i.e., $- \sqrt 2 < \sin x - \cos x < \sqrt 2$ , where, $x$ is just an arbitrary angle which belongs to the domain of the function. The main point here is that angles of both the $\sin$ function and $\cos$ function must be the same. Hence, it is also true for $- \sqrt 2 < \sin 2x - \cos 2x < \sqrt 2$ , as angles of both the functions are the same.
One must have the basic properties of solving inequalities.
If we multiply, divide, add or subtract the same number on each side, then the equation does not change.