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Question: The solution of the equation \({\sin ^{ - 1}}\left( {\tan \dfrac{\pi }{4}} \right) - {\sin ^{ - 1}}\...

The solution of the equation sin⁡−1(tan⁡π4)−sin⁡−1(3x)−π6=0{\sin ^{ - 1}}\left( {\tan \dfrac{\pi }{4}} \right) - {\sin ^{ - 1}}\left( {\sqrt {\dfrac{3}{x}} } \right) - \dfrac{\pi }{6} = 0 is given by x=x =
A) 11
B) 22
C) 33
D) 44

Explanation

Solution

In this question we have been given an equation sin⁡−1(tan⁡π4)−sin⁡−1(3x)−π6=0{\sin ^{ - 1}}\left( {\tan \dfrac{\pi }{4}} \right) - {\sin ^{ - 1}}\left( {\sqrt {\dfrac{3}{x}} } \right) - \dfrac{\pi }{6} = 0. We can see that there are trigonometric ratios in these types of equations, so we will use the trigonometric identities to solve this question.
We will try to eliminate the sine from the equation, so we will first use the values. We know that the value of tan⁡π4\tan \dfrac{\pi }{4} can be also be written as tan⁡1804=tan⁡45∘\tan \dfrac{{180}}{4} = \tan {45^ \circ }
And we know that the value of tan⁡45∘=1\tan {45^ \circ } = 1 . So we can say that tan⁡π4=1\tan \dfrac{\pi }{4} = 1 .

Complete step by step solution:
Here we have sin⁡−1(tan⁡π4)−sin⁡−1(3x)−π6=0{\sin ^{ - 1}}\left( {\tan \dfrac{\pi }{4}} \right) - {\sin ^{ - 1}}\left( {\sqrt {\dfrac{3}{x}} } \right) - \dfrac{\pi }{6} = 0
We know that the value of
tan⁡π4=1\tan \dfrac{\pi }{4} = 1
So we can write the equation as
sin⁡−1(1)−sin⁡−1(3x)−π6=0{\sin ^{ - 1}}\left( 1 \right) - {\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{{\sqrt x }}} \right) - \dfrac{\pi }{6} = 0 .
We should keep in mind that sin⁡−1(1){\sin ^{ - 1}}(1) is equal to the angle whose sine value is 11 .
And we know that the value of sin⁡90∘=1\sin {90^ \circ } = 1
So we can say that the inverse of sin⁡−1(1){\sin ^{ - 1}}(1) is 90∘{90^ \circ }.
We can also write 9090 as π2\dfrac{\pi }{2}
Now we can write the expression as
π2−sin⁡−1(3x)−π6=0\dfrac{\pi }{2} - {\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{{\sqrt x }}} \right) - \dfrac{\pi }{6} = 0 .
We will take the similar terms to the other sides of the equation and we have :
−sin⁡−1(3x)=π6−π2- {\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{{\sqrt x }}} \right) = \dfrac{\pi }{6} - \dfrac{\pi }{2} .
On simplifying we have :
−sin⁡−1(3x)=π−3π6⇒−sin⁡−1(3x)=−2π6- {\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{{\sqrt x }}} \right) = \dfrac{{\pi - 3\pi }}{6} \Rightarrow - {\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{{\sqrt x }}} \right) = - \dfrac{{2\pi }}{6} .
After eliminating the negative sign, the equation can also be written as :
sin⁡−1(3x)=π3{\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{{\sqrt x }}} \right) = \dfrac{\pi }{3}
By taking sin⁡\sin on both the sides of the equation we have:
sin⁡(sin⁡−1(3x))=sin⁡π3\sin \left( {{{\sin }^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{{\sqrt x }}} \right)} \right) = \sin \dfrac{\pi }{3}
We know the trigonometric identity that
sin⁡(sin⁡−1θ)=θ\sin \left( {{{\sin }^{ - 1}}\theta } \right) = \theta
So we can write
sin⁡(sin⁡−1(3x))=3x\sin \left( {{{\sin }^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{{\sqrt x }}} \right)} \right) = \dfrac{{\sqrt 3 }}{x}.
And the value of sin⁡π3=32\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}
By putting the value back in the equation we have :
3x=32\dfrac{{\sqrt 3 }}{{\sqrt x }} = \dfrac{{\sqrt 3 }}{2}
By cross multiplication we have;
x=233⇒x=2\sqrt x = \dfrac{{2\sqrt 3 }}{{\sqrt 3 }} \Rightarrow \sqrt x = 2
We can square both the sides to eliminate the under root sign i.e.
(x)2=22{\left( {\sqrt x } \right)^2} = {2^2}
So it gives us value x=4x = 4
Hence the correct option is (D) 44.

Note:
We should note that
π=180∘\pi = {180^ \circ } .
So in the above solution, we can write π3\dfrac{\pi }{3} as
1803=60\dfrac{{180}}{3} = 60
It gives us
sin⁡60∘\sin {60^ \circ }
And we know the trigonometric value that
sin⁡60∘=32\sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2} .
So it gives us the value of
sin⁡π3=32\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2} .