Question

# Solve the equation ${\left( {\sin \theta } \right)^3}\left( {\cos \theta } \right) - {\left( {\cos \theta } \right)^3}\left( {\sin \theta } \right) = \dfrac{1}{4}$ for the value of $\theta$.

Hint- Here, we will be simplifying the LHS of the given equation by using the formulas which are $\sin 2\theta = 2\left( {\sin \theta } \right)\left( {\cos \theta } \right)$ and $\cos 2\theta = \left[ {{{\left( {\cos \theta } \right)}^2} - {{\left( {\sin \theta } \right)}^2}} \right]$. Then, with the help of the condition i.e., when $\sin x = - 1$ then $x = 2n\pi - \dfrac{\pi }{2}$ we will find the required values of $\theta$.

The given equation is ${\left( {\sin \theta } \right)^3}\left( {\cos \theta } \right) - {\left( {\cos \theta } \right)^3}\left( {\sin \theta } \right) = \dfrac{1}{4}$.
This equation can be simplified as under
$\Rightarrow \left( {\sin \theta } \right)\left( {\cos \theta } \right)\left[ {{{\left( {\sin \theta } \right)}^2} - {{\left( {\cos \theta } \right)}^2}} \right] = \dfrac{1}{4} \\ \Rightarrow - 4\left( {\sin \theta } \right)\left( {\cos \theta } \right)\left[ {{{\left( {\cos \theta } \right)}^2} - {{\left( {\sin \theta } \right)}^2}} \right] = 1 \\ \Rightarrow - 2\left[ {2\left( {\sin \theta } \right)\left( {\cos \theta } \right)} \right]\left[ {{{\left( {\cos \theta } \right)}^2} - {{\left( {\sin \theta } \right)}^2}} \right] = 1{\text{ }} \to {\text{(1)}} \\$
As we know that $\sin 2\theta = 2\left( {\sin \theta } \right)\left( {\cos \theta } \right)$ and $\cos 2\theta = \left[ {{{\left( {\cos \theta } \right)}^2} - {{\left( {\sin \theta } \right)}^2}} \right]$
Using the above formulas, equation (1) becomes,
$\Rightarrow - 2\left[ {\sin 2\theta } \right]\left[ {\cos 2\theta } \right] = 1 \\ \Rightarrow - \left[ {2\left( {\sin 2\theta } \right)\left( {\cos 2\theta } \right)} \right] = 1{\text{ }} \to {\text{(2)}} \\$
In the formula $\sin 2\theta = 2\left( {\sin \theta } \right)\left( {\cos \theta } \right)$ replace $\theta$ by $2\theta$, we get
$\Rightarrow \sin 2\left( {2\theta } \right) = 2\left( {\sin 2\theta } \right)\left( {\cos 2\theta } \right) \\ \Rightarrow \sin 4\theta = 2\left( {\sin 2\theta } \right)\left( {\cos 2\theta } \right){\text{ }} \to {\text{(3)}} \\$
Using equation (3) in equation (2), we get
$\Rightarrow - \left[ {\sin 4\theta } \right] = 1 \\ \Rightarrow \sin 4\theta = - 1 \\$
Also we know that when $\sin x = - 1$ then $x = 2n\pi - \dfrac{\pi }{2}$ where $n \in Z$ (Z is set of integers)
Replacing x in the above condition by $4\theta$, we get
$4\theta = 2n\pi - \dfrac{\pi }{2} \\ \Rightarrow \theta = \dfrac{{2n\pi }}{4} - \dfrac{\pi }{{2 \times 4}} \\ \Rightarrow \theta = \dfrac{{n\pi }}{2} - \dfrac{\pi }{8}{\text{ where }}n \in Z \\$

Note- In these types of problems, we convert the trigonometric functions of smaller angle (i.e., $\theta$ in this case) in the given equation to the trigonometric functions of larger angle (i.e. $4\theta$ in this case). Also, in this problem $\sin 4\theta = - 1$ is coming after simplification which means the values of $4\theta$ which are possible are $2n\pi - \dfrac{\pi }{2}$ where n belongs to the set of integers.