Question

${\cos ^3}\alpha + {\cos ^3}\left( {{{120}^ \circ } + \alpha } \right) + {\cos ^3}\left( {{{120}^ \circ } - \alpha } \right) = \dfrac{{3\sqrt 3 }}{4}$

Then the general solution $\alpha$ is

1.$\phi$

2.$2n\pi \pm \dfrac{\pi }{3},\forall n \in Z$

3.$(2n + 1)\dfrac{\pi }{2},\forall n \in Z$

4.$n\pi ,\forall n \in Z$

Answer 1

For a question like this we approach the solution by simplifying solving the one side expressions as here we’ll simplify the left-hand side using some of the trigonometric formulas like
$co{s^3}x = \dfrac{{cos3x + 3cosx}}{4}$
$cos\left( {A + B} \right) = \cos A\cos B - \sin A\sin B$ and
$cos\left( {A - B} \right) = \cos A\cos B + \sin A\sin B$
We simplify in such a manner that we can easily compare it with the other side expression or in such a manner that we can get an equation we can easily get the result of like a quadratic equation or of linear form.

Complete step-by-step answer:
Given data: ${\cos ^3}\alpha + {\cos ^3}\left( {{{120}^ \circ } + \alpha } \right) + {\cos ^3}\left( {{{120}^ \circ } - \alpha } \right) = \dfrac{{3\sqrt 3 }}{4}$
Solving for the given equation i.e. ${\cos ^3}\alpha + {\cos ^3}\left( {{{120}^ \circ } + \alpha } \right) + {\cos ^3}\left( {{{120}^ \circ } - \alpha } \right) = \dfrac{{3\sqrt 3 }}{4}$
We know that $cos3x = 4co{s^3}x - 3cosx$ from this we have the formula for $co{s^3}x$i.e.
$co{s^3}x = \dfrac{{cos3x + 3cosx}}{4}$, using this formula we get
$\Rightarrow \dfrac{{cos3\alpha + 3cos\alpha }}{4} + \dfrac{{cos3\left( {{{120}^ \circ } + \alpha } \right) + 3cos\left( {{{120}^ \circ } + \alpha } \right)}}{4} + \dfrac{{cos3\left( {{{120}^ \circ } - \alpha } \right) + 3cos\left( {{{120}^ \circ } - \alpha } \right)}}{4} = \dfrac{{3\sqrt 3 }}{4}$
Now, simplifying the brackets we get,
$\Rightarrow \dfrac{{cos3\alpha + 3cos\alpha }}{4} + \dfrac{{cos\left( {{{360}^ \circ } + 3\alpha } \right) + 3cos\left( {{{120}^ \circ } + \alpha } \right)}}{4} + \dfrac{{cos\left( {{{360}^ \circ } - 3\alpha } \right) + 3cos\left( {{{120}^ \circ } - \alpha } \right)}}{4} = \dfrac{{3\sqrt 3 }}{4}$
We know that $cos\left( {{{360}^ \circ } - x} \right) = \cos x$, we get

$\Rightarrow \dfrac{{cos3\alpha + 3cos\alpha }}{4} + \dfrac{{cos3\alpha + 3cos\left( {{{120}^ \circ } + \alpha } \right)}}{4} + \dfrac{{cos3\alpha + 3cos\left( {{{120}^ \circ } - \alpha } \right)}}{4} = \dfrac{{3\sqrt 3 }}{4}$
Multiplying both the equation with 4, we get
$\Rightarrow cos3\alpha + 3cos\alpha + cos3\alpha + 3cos\left( {{{120}^ \circ } + \alpha } \right) + cos3\alpha + 3cos\left( {{{120}^ \circ } - \alpha } \right) = 3\sqrt 3$
Simplifying the like terms we get,
$\Rightarrow 3cos3\alpha + 3cos\alpha + 3cos\left( {{{120}^ \circ } + \alpha } \right) + 3cos\left( {{{120}^ \circ } - \alpha } \right) = 3\sqrt 3$
Dividing both the sides by 3, we get
$\Rightarrow cos3\alpha + cos\alpha + cos\left( {{{120}^ \circ } + \alpha } \right) + cos\left( {{{120}^ \circ } - \alpha } \right) = \sqrt 3$
Now using formulas $cos\left( {A + B} \right) = \cos A\cos B - \sin A\sin B$ and $cos\left( {A - B} \right) = \cos A\cos B + \sin A\sin B$
$\Rightarrow cos3\alpha + cos\alpha + \cos {120^ \circ }\cos \alpha - \sin {120^ \circ }\sin \alpha + \cos {120^ \circ }\cos \alpha + \sin {120^ \circ }\sin \alpha = \sqrt 3$
On simplifying we get,
$\Rightarrow cos3\alpha + cos\alpha + 2\cos {120^ \circ }\cos \alpha = \sqrt 3$
Now, substituting $\cos {120^ \circ } = - \dfrac{1}{2}$
$\Rightarrow 2\cos 2\alpha \cos \alpha - 2 \times \dfrac{1}{2}\cos \alpha = \sqrt 3$
On simplifying we get,
$\Rightarrow cos3\alpha + cos\alpha - \cos \alpha = \sqrt 3$
We get, $cos3\alpha = \sqrt 3$
We know that the cosine function can only give an answer in $[ - 1,1]$
i.e. the range of the cosine function is $[ - 1,1]$ and here it is equal to $\sqrt 3$ which is greater than 1
therefore the given equation has no solution.
Option(A) is correct

Note: In questions of trigonometric functions, our priority should be using trigonometric properties over the algebraic formulas, like in this question if we have used the formula for the sum of the cube of three terms in place of the cube of the cosine function the problem might have gotten more complex, so try using the trigonometric function if possible.