Question

Solve the question and please explain each step properly. ​

Answer 1

Answer:

First, let us convert all the given trigonometric values in terms of sin A and cos A. Hence we get:

$\implies \dfrac{Sin^3A}{Cos^3A} + \dfrac{Cos^3A}{Sin^3A} = 12 + 8 \times ( \dfrac{1}{Sin^3\:2A} )\\\\\\\implies \dfrac{Sin^3A}{Cos^3A} + \dfrac{Cos^3A}{Sin^3A} = 12 + \dfrac{8}{(2.\:SinA.\:CosA)^3}\\\\\\\text{Taking LCM on both sides we get:}\\\\\\\implies \dfrac{ (Sin^3A \times Sin^3A) + (Cos^3A \times Cos^3A)}{(Sin^3A.\:Cos^3A)} = 12 + \dfrac{8}{8.\:Sin^3A.\:Cos^3A}\\\\\\\implies \dfrac{ Sin^6A+Cos^6A}{Sin^3A.\:Cos^3A} = \dfrac{12.\:Sin^3A.\:Cos^3A + 1 }{ Sin^3A.\:Cos^3A}$

$\text{Cancelling the denominator on both sides, we get:}\\\\\\ \implies Sin^6A + Cos^6A = 12.\:Sin^3A.\:Cos^3A + 1\ \:\:...(i)\\\\\\\text{According to an identity, we know that:} \\\\\\ \implies Cos^6A + Sin^6A = 1 - 3.\:Sin^2A.\:Cos^2A \:\: ...(ii)\\\\\\\text{Substituting (ii) in (i) we get:}\\\\\\\implies 1 - 3.\:Sin^2A.\:Cos^2A = 12.\:Sin^3A.\:Cos^3A + 1\\\\\\\implies 12.\:Sin^3A.\:Cos^3A + 3.\:Sin^2A.\:Cos^2A = 0$

$\implies (3.\:Sin^2A.\:Cos^2A)\:\: [\:4.\:SinA.\:CosA + 1\:] = 0\\\\\\\text{Simplifying it further, we get:}\\\\\\\implies (3.\:Sin^2A.\:Cos^2A)\:\: [\:2.\;Sin.\:2A + 1\:] = 0\\\\\\\implies 2.\;Sin.\:2A + 1 = 0\\\\\\\implies 2.\:Sin.\:2A = -1\\\\\\\implies Sin.\:2A = \dfrac{-1}{2}$

Writing the solution in terms of general solution, we get:

$\implies 2A = sin^{-1}(\dfrac{-1}{2}) = (\pi + \dfrac{\pi}{6}), (\pi + \dfrac{5\pi}{6}) \\\\\\\text{The general solution for the above solution is:} \\\\\\\implies 2A = (2n+1) \pi + \dfrac{ \pi}{6} \:\:and\:\: (2n+1) \pi + \dfrac{5 \pi}{6}\\\\\\\implies A = \boxed{(2n + 1 ) \dfrac{ \pi}{2} + \dfrac{\pi}{12}} \:\:and\:\: \boxed{(2n+1)\pi + \dfrac{5\pi}{12}}$

Since the interval mentioned is [ 0, 2π ], the possible values of 'n' are 0 & 1. Hence substituting the value of 'n', we get:

$\text{ For n} = 0,\\\\\implies A = [ 2(0) + 1 ]\dfrac{\pi}{2} + \dfrac{\pi}{12} \:\: and \:\: [2(0) + 1]\dfrac{\pi}{2} + \dfrac{5\pi}{12}\\\\\\\implies A = \boxed{\bf{\dfrac{7\pi}{12}}} \:\:and\:\: \boxed{\bf{\dfrac{11 \pi}{12}}}$

$\text{ For n} = 1,\\\\\implies A = [ 2(1) + 1 ]\dfrac{\pi}{2} + \dfrac{\pi}{12} \:\: and \:\: [2(1) + 1]\dfrac{\pi}{2} + \dfrac{5\pi}{12}\\\\\\\implies A = (\dfrac{3 \pi}{2} + \dfrac{\pi}{12}) \:\: and \:\: (\dfrac{3\pi}{2} + \dfrac{5\pi}{12}) \\\\\\\implies A = \boxed{\bf{\dfrac{19\pi}{12}}} \:\:and\:\: \boxed{\bf{\dfrac{23 \pi}{12}}}$

Hence the number of values are 4.