Question

Solve the equation:
$\cos 3x.{\cos ^3}x + \sin 3x.{\sin ^3}x = 0$
$
  (a){\text{ }}\left( {2n + 1} \right)\dfrac{\pi }{4},{\text{ n}} \in {\text{I}} \\
  {\text{(b) }}\left( {2n + 1} \right)\dfrac{\pi }{2},{\text{ n}} \in {\text{I}} \\
  {\text{(c) }}\left( {2n + 1} \right)\dfrac{\pi }{3},{\text{ n}} \in {\text{I}} \\
  {\text{(d) }}\left( {2n + 1} \right)\dfrac{\pi }{6},{\text{ n}} \in {\text{I}} \\
 $

Answer 1

Hint – In this question use the formula that $\cos 3x = 4{\cos ^3}x - 3\cos x,\sin 3x = 3\sin x - 4{\sin ^3}x$. On simplification and application of basic algebraic identity$\left[ {{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)} \right],\left[ {{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)} \right]$, will help manipulating the equation to the final result.

Complete step-by-step answer:
Given trigonometric equation is
$\cos 3x.{\cos ^3}x + \sin 3x.{\sin ^3}x = 0$
Now as we know that $\cos 3x = 4{\cos ^3}x - 3\cos x,\sin 3x = 3\sin x - 4{\sin ^3}x$ so use this property in above equation that is
$ \Rightarrow \left( {4{{\cos }^3}x - 3\cos x} \right).{\cos ^3}x + \left( {3\sin x - 4{{\sin }^3}x} \right).{\sin ^3}x = 0$
Now simplify this equation we have,
$ \Rightarrow 4{\cos ^6}x - 3{\cos ^4}x + 3{\sin ^4}x - 4{\sin ^6}x = 0$
$ \Rightarrow 4\left( {{{\cos }^6}x - {{\sin }^6}x} \right) - 3\left( {{{\cos }^4}x - {{\sin }^4}x} \right) = 0$
$ \Rightarrow 4\left[ {{{\left( {{{\cos }^2}x} \right)}^3} - {{\left( {{{\sin }^2}x} \right)}^3}} \right] - 3\left[ {{{\left( {{{\cos }^2}x} \right)}^2} - {{\left( {{{\sin }^2}x} \right)}^2}} \right] = 0$
Now use the property that $\left[ {{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)} \right],\left[ {{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)} \right]$ so we have,
$ \Rightarrow 4\left[ {\left( {{{\cos }^2}x} \right) - \left( {{{\sin }^2}x} \right)} \right]\left[ {{{\cos }^4}x + {{\sin }^4}x + {{\cos }^2}x{{\sin }^2}x} \right] - 3\left[ {\left( {{{\cos }^2}x} \right) - \left( {{{\sin }^2}x} \right)} \right]\left[ {{{\cos }^2}x + {{\sin }^2}x} \right] = 0$
Now as we know that ${\sin ^2}x + {\cos ^2}x = 1$ so we have,
$ \Rightarrow \left( {{{\cos }^2}x - {{\sin }^2}x} \right)\left[ {4\left( {{{\cos }^4}x + {{\sin }^4}x + {{\cos }^2}x{{\sin }^2}x} \right) - 3} \right] = 0$
Now add and subtract by ${\cos ^2}x{\sin ^2}x$ in the term ${\cos ^4}x + {\sin ^4}x + {\cos ^2}x{\sin ^2}x$ we have
$ \Rightarrow \left( {\cos 2x} \right)\left[ {4\left( {{{\cos }^4}x + {{\sin }^4}x + 2{{\cos }^2}x{{\sin }^2}x - {{\cos }^2}x{{\sin }^2}x} \right) - 3} \right] = 0$, $\left[ {\because {{\cos }^2}x - {{\sin }^2}x = \cos 2x} \right]$
Now use the property ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ we have,
$ \Rightarrow \left( {\cos 2x} \right)\left[ {4{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^2} - 4{{\cos }^2}x{{\sin }^2}x - 3} \right] = 0$
$ \Rightarrow \left( {\cos 2x} \right)\left[ {4 - 4{{\cos }^2}x{{\sin }^2}x - 3} \right] = 0$
$ \Rightarrow \left( {\cos 2x} \right)\left[ {1 - {{\left( {2\cos x\sin x} \right)}^2}} \right] = 0$
$ \Rightarrow \left( {\cos 2x} \right)\left[ {1 - {{\sin }^2}2x} \right] = 0$, $\left[ {\because \sin 2x = 2\sin x\cos x} \right]$
$ \Rightarrow \left( {\cos 2x} \right)\left[ {{{\cos }^2}2x} \right] = 0$, $\left[ {\because 1 - {{\sin }^2}2x = {{\cos }^2}2x} \right]$
$ \Rightarrow {\cos ^3}2x = 0$
$ \Rightarrow \cos 2x = 0 = \cos \left[ {\left( {2n + 1} \right)\dfrac{\pi }{2}} \right]$ Where, n = 0, 1, 2, 3...
Now on comparing we have,
$2x = \left( {2n + 1} \right)\dfrac{\pi }{2}$
$ \Rightarrow x = \left( {2n + 1} \right)\dfrac{\pi }{4}$, $n \in I$
So this is the required solution.
Hence option (A) is the correct answer.

Note – It is always advisable to remember basic trigonometric identity like $\sin 2x = 2\sin x\cos x$, $1 - {\sin ^2}2x = {\cos ^2}2x$ and others like $\cos 2x = {\cos ^2}x - {\sin ^2}x$, $1 + {\tan ^2}x = {\sec ^2}x$ etc. helps solving problems of this kind. The verification of $\cos 4\theta = 0 = \cos \left[ {\left( {2n + 1} \right)\dfrac{\pi }{2}} \right]$ can be provided by the fact that n in this case varies form n = 0, 1, 2... that is set of whole numbers. So if n=0, then we will have $\cos \dfrac{\pi }{2}$ which eventually will be zero. Now if we substitute 1 in place of n we get $\cos \dfrac{{3\pi }}{2}$ which is again zero. Thus $\cos \left[ {\left( {2n + 1} \right)\dfrac{\pi }{2}} \right]$ is the general value for any $\cos \theta = 0$.