Question

The value of x satisfying $\dfrac{d}{{dx}}\left[ {\cos x + a} \right] = \sin 2x$ are
A). $\dfrac{{3n\pi }}{2}$
B). $\dfrac{{3n\pi }}{4}$
C). $\dfrac{{2n\pi }}{3}$
D). None of these

Answer 1

Here in the question $\dfrac{d}{{dx}}$ is nothing but the concept of derivative of the function so given with respect to x. We will directly go for the derivative of the function given. Then we will use formulas for the angles so found. With the help of that we can equate the angles and get the value of x.

Complete step-by-step solution:
Given that,
$\dfrac{d}{{dx}}\left[ {\cos x + a} \right] = \sin 2x$
We know that the derivative of $\cos x$ is $- \sin x$ and that of a constant is zero.
$- \sin x = \sin 2x$
Now transposing the sin function we get,
$\sin 2x + \sin x = 0$
Now using the formula of $\sin A + \sin B$ we get,
$2\sin \left( {\dfrac{{2x + x}}{2}} \right).\cos \left( {\dfrac{{2x - x}}{2}} \right) = 0$
Adding and subtracting the angle,
$2\sin \left( {\dfrac{{3x}}{2}} \right).\cos \left( {\dfrac{x}{2}} \right) = 0$
This can be simplified as,
$\sin \left( {\dfrac{{3x}}{2}} \right).\cos \left( {\dfrac{x}{2}} \right) = 0$
Now equating the angles separately,
$\sin \left( {\dfrac{{3x}}{2}} \right) = 0or\cos \left( {\dfrac{x}{2}} \right) = 0$
Then,
$\dfrac{{3x}}{2} = \pi or\dfrac{x}{2} = \dfrac{\pi }{2}$
So x equals to,
$x = \dfrac{{2n\pi }}{3}$
This is the correct answer.
So option C is correct.

Note: Note that, we will find the derivative of LHS only and not of RHS. Because it is not instructed to find the derivative of both sides. So be careful. Also note that the angle is not fixed; it is in terms of n so no need to find any particular angle.