Question

General solution of the equation ${\sin ^3}\theta \cos \theta - \sin \theta {\cos ^3}\theta = \dfrac{1}{4}$ is
A. $n\left( {\dfrac{\pi }{4}} \right) + \left( {\dfrac{\pi }{8}} \right)$
B. $n\left( {\dfrac{\pi }{2}} \right) \pm \left( {\dfrac{\pi }{8}} \right)$
C. $n\left( {\dfrac{\pi }{4}} \right) + {\left( { - 1} \right)^{n + 1}}\left( {\dfrac{\pi }{8}} \right)$
D. $n\left( {\dfrac{\pi }{4}} \right) + {\left( { - 1} \right)^n}\left( {\dfrac{\pi }{8}} \right)$

Answer 1

Hint:The given question involves solving a trigonometric equation and finding the general value of angle $x$ that satisfies the given equation. There can be various methods to solve a specific trigonometric equation. For solving such questions, we need to have knowledge of basic trigonometric formulae and identities. We must know the double angle formulae of sine and cosine to tackle the given problem.

Complete step by step answer:
In the given problem, we have to solve the trigonometric equation ${\sin ^3}\theta \cos \theta - \sin \theta {\cos ^3}\theta = \dfrac{1}{4}$ and find the general solutions of x that satisfy the given equation.
So, in order to solve the given trigonometric equation ${\sin ^3}\theta \cos \theta - \sin \theta {\cos ^3}\theta = \dfrac{1}{4}$ , we should first take the common terms outside. So, we get,
$ \Rightarrow \sin \theta \cos \theta \left( {{{\sin }^2}\theta - {{\cos }^2}\theta } \right) = \dfrac{1}{4}$
Now, we know that the double angle formula for cosine is $\cos 2x = {\cos ^2}x - {\sin ^2}x$. Hence, we get,
$ \Rightarrow \sin \theta \cos \theta \left( { - \cos 2\theta } \right) = \dfrac{1}{4}$
Multiplying and dividing the equation by $2$, we get,
$ \Rightarrow \dfrac{{2\sin \theta \cos \theta \left( { - \cos 2\theta } \right)}}{2} = \dfrac{1}{4}$
Now, we know that the double angle formula for sine is $\sin 2x = 2\sin x\cos x$. Hence, we get,
$ \Rightarrow \dfrac{{ - \sin 2\theta \cos 2\theta }}{2} = \dfrac{1}{4}$
Again, multiplying and dividing the equation by $2$, we get,
$ \Rightarrow \dfrac{{ - 2\sin 2\theta \cos 2\theta }}{4} = \dfrac{1}{4}$
Using the double angle formula for sine is $\sin 2x = 2\sin x\cos x$, we get,
$ \Rightarrow \dfrac{{ - \sin 4\theta }}{4} = \dfrac{1}{4}$
Now, cross multiplying the terms, we get,
$ \Rightarrow \sin 4\theta = - 1$
So, we know that the value of $\sin \left( { - \dfrac{\pi }{2}} \right)$ is $\left( { - 1} \right)$. So, we get,
$ \Rightarrow \sin 4\theta = \sin \left( { - \dfrac{\pi }{2}} \right)$
Now, we know that the general solution of the equation $\sin x = \sin \alpha $ is of the form $x = n\pi + {\left( { - 1} \right)^n}\alpha $.
So, the general solution of the trigonometric equation ${\sin ^3}\theta \cos \theta - \sin \theta {\cos ^3}\theta = \dfrac{1}{4}$ is $4\theta = n\pi + {\left( { - 1} \right)^n}\left( { - \dfrac{\pi }{2}} \right)$
$ \Rightarrow \theta = n\left( {\dfrac{\pi }{4}} \right) + {\left( { - 1} \right)^n}\left( { - \dfrac{\pi }{8}} \right)$
Simplifying the expression to match the options, we get,
$ \Rightarrow \theta = n\left( {\dfrac{\pi }{4}} \right) + {\left( { - 1} \right)^{n + 1}}\left( {\dfrac{\pi }{8}} \right)$
So, we get the general solution of the trigonometric equation ${\sin ^3}\theta \cos \theta - \sin \theta {\cos ^3}\theta = \dfrac{1}{4}$ as $\theta = n\left( {\dfrac{\pi }{4}} \right) + {\left( { - 1} \right)^{n + 1}}\left( {\dfrac{\pi }{8}} \right)$.

So, option C is correct.

Note:There are two types of solution of a trigonometric equation: General solution and principal solution. Principal solutions refer to the solution of the equation that lies in the range $\left[ {0,2\pi } \right]$. General solution represents all the possible values of the variable x in the trigonometric equation with the use of a parameter. We can get all the possible values of the variable x by substituting in the integral values of the parameter.