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Question: Find real value of x, if \( {\cos ^{ - 1}}\left( {\sqrt 6 x} \right) + {\cos ^{ - 1}}\left( {3\s...

Find real value of x, if
cos⁡−1(6x)+cos⁡−1(33x2)=π2{\cos ^{ - 1}}\left( {\sqrt 6 x} \right) + {\cos ^{ - 1}}\left( {3\sqrt 3 {x^2}} \right) = \dfrac{\pi }{2}
A.  x=±13 B.   x=±12 C.  x=±13 D.  x=±12  A.\,\,x = \pm \dfrac{1}{3} \\\ B.\,\,\,x = \pm \dfrac{1}{2} \\\ C.\,\,x = \pm \dfrac{1}{{\sqrt 3 }} \\\ D.\,\,x = \pm \dfrac{1}{{\sqrt 2 }} \\\

Explanation

Solution

Hint : In this type of problem we shift either of cosine term to right hand side then using property to write or convert it in sine trigonometric term and then suing sine conversion formula to write sine trigonometric term to cosine trigonometric term and then ceiling cosine function from both side and then solving equation so formed to find value of ‘x’ and hence solution of given problem.
Formulas used: π2−cos⁡−1A=sin⁡−1A\dfrac{\pi }{2} - {\cos ^{ - 1}}A = {\sin ^{ - 1}}A , sin⁡−1θ=cos⁡−11−θ2{\sin ^{ - 1}}\theta = {\cos ^{ - 1}}\sqrt {1 - {\theta ^2}}

Complete step by step solution:
Given equation cos⁡−1(6x)+cos⁡−1(33x2)=π2{\cos ^{ - 1}}\left( {\sqrt 6 x} \right) + {\cos ^{ - 1}}\left( {3\sqrt 3 {x^2}} \right) = \dfrac{\pi }{2}
We can write above equation as:
cos⁡−1(33x2)=π2−cos⁡−1(6x){\cos ^{ - 1}}\left( {3\sqrt 3 {x^2}} \right) = \dfrac{\pi }{2} - {\cos ^{ - 1}}\left( {\sqrt 6 x} \right)
Also, we know that π2−cos⁡−1(θ)=sin⁡−1(θ)\dfrac{\pi }{2} - {\cos ^{ - 1}}(\theta ) = {\sin ^{ - 1}}(\theta )
Using the above mentioned identity in the above formed equation. We have,
cos⁡−1(33x2)=sin⁡−1(6x){\cos ^{ - 1}}\left( {3\sqrt 3 {x^2}} \right) = {\sin ^{ - 1}}\left( {\sqrt 6 x} \right)
Also, we know that sin⁡−1θ=cos⁡−11−θ2{\sin ^{ - 1}}\theta = {\cos ^{ - 1}}\sqrt {1 - {\theta ^2}}
Using above trigonometric identity in above formed equation. We have,
cos⁡−1(33x2)=cos⁡−1[1−(6x)2] ⇒33x2=1−6x2   {\cos ^{ - 1}}\left( {3\sqrt 3 {x^2}} \right) = {\cos ^{ - 1}}\left[ {\sqrt {1 - {{\left( {\sqrt {6x} } \right)}^2}} } \right] \\\ \Rightarrow 3\sqrt 3 {x^2} = \sqrt {1 - 6{x^2}} \;
Squaring both sides to solve it.
(33x2)2=(1−6x2)2 ⇒9(3x4)=1−6x2 ⇒27x4+6x2−1=0   {\left( {3\sqrt 3 {x^2}} \right)^2} = {\left( {\sqrt {1 - 6{x^2}} } \right)^2} \\\ \Rightarrow 9\left( {3{x^4}} \right) = 1 - 6{x^2} \\\ \Rightarrow 27{x^4} + 6{x^2} - 1 = 0 \;
Taking x2=A{x^2} = A above equation becomes.
27(x2)2+6(x2)−1=0 ⇒27A2+6A−1=0   27{({x^2})^2} + 6({x^2}) - 1 = 0 \\\ \Rightarrow 27{A^2} + 6A - 1 = 0 \;
Solving the above formed quadratic equation by middle term splitting method.
27A2+9A−3A−1=0 ⇒9A(3A+1)−(3A+1)=0 ⇒(3A+1)(9A−1)=0 ⇒3A+1=0   or   9A−1=0 ⇒A=−13   or   A=19  27{A^2} + 9A - 3A - 1 = 0 \\\ \Rightarrow 9A\left( {3A + 1} \right) - \left( {3A + 1} \right) = 0 \\\ \Rightarrow \left( {3A + 1} \right)\left( {9A - 1} \right) = 0 \\\ \Rightarrow 3A + 1 = 0\,\,\,or\,\,\,9A - 1 = 0 \\\ \Rightarrow A = - \dfrac{1}{3}\,\,\,or\,\,\,A = \dfrac{1}{9} \\\
Now, substituting value of A in above. We have,
x2=19    or  x2=−13{x^2} = \dfrac{1}{9}\,\,\,\,or\,\,{x^2} = - \dfrac{1}{3}
But x2=−13{x^2} = - \dfrac{1}{3} is clearly not possible.
Therefore, considering

x2=19 ⇒x=19 ⇒x=±13  {x^2} = \dfrac{1}{9} \\\ \Rightarrow x = \sqrt {\dfrac{1}{9}} \\\ \Rightarrow x = \pm \dfrac{1}{3} \;

Therefore, from above we see that required value of ‘x’ is ±13\pm \dfrac{1}{3}
So, the correct answer is “Option C”.

Note : We can also find the value of ‘x’ in other ways. In this we directly apply trigonometric identity of cos⁡−1A+cos⁡−1B=cos⁡−1(AB−1−A21−B2){\cos ^{ - 1}}A + {\cos ^{ - 1}}B = {\cos ^{ - 1}}\left( {AB - \sqrt {1 - {A^2}} \sqrt {1 - {B^2}} } \right) on left hand side and then shifting cos⁡−1{\cos ^{ - 1}} to right hand side to form an equation. On solving this equation by squaring and simplifying we can get the value of ‘x’ or we can say the required solution to the given problem.