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Question: If \(\cos \left( {2{{\sin }^{ - 1}}x} \right) = \dfrac{1}{9}\), then \(x\) is equal to A) Only \(\...

If cos⁡(2sin⁡−1x)=19\cos \left( {2{{\sin }^{ - 1}}x} \right) = \dfrac{1}{9}, then xx is equal to
A) Only 23\dfrac{2}{3}
B) Only −23 - \dfrac{2}{3}
C) 23,−23\dfrac{2}{3}, - \dfrac{2}{3}
D) 13\dfrac{1}{3}

Explanation

Solution

First, assume sin⁡−1x{\sin ^{ - 1}}x as another variable. Then use the fact that cos⁡2A=1−2sin⁡2A\cos 2A = 1 - 2{\sin ^2}A. Use the fact to convert the term on the left side in terms of the sign. After that, move the constant part to the other side and simplify them and cancel out the negative sign from both sides. Then take the square root. After that substitute back sin⁡−1x{\sin ^{ - 1}}x in place of the variable and use the property sin⁡(sin⁡−1x)=x\sin \left( {{{\sin }^{ - 1}}x} \right) = x to remove the sine part to get the value of x.

Complete step-by-step solution:
Given:- cos⁡(2sin⁡−1x)=19\cos \left( {2{{\sin }^{ - 1}}x} \right) = \dfrac{1}{9}
The inverse trigonometric functions perform the opposite operation of the trigonometric functions such as sine, cosine, tangent, cosecant, secant, and cotangent. We know that trig functions are especially applicable to the right-angle triangle.
Let us assume y=sin⁡−1xy = {\sin ^{ - 1}}x.
Then substitute in the expression,
⇒cos⁡2y=19\Rightarrow \cos 2y = \dfrac{1}{9}
As we know that,
cos⁡2A=1−2sin⁡2A\cos 2A = 1 - 2{\sin ^2}A
So, expand cos⁡2y\cos 2y using this formula,
⇒1−2sin⁡2y=19\Rightarrow 1 - 2{\sin ^2}y = \dfrac{1}{9}
Move the constant part on one side,
⇒−2sin⁡2y=19−1\Rightarrow - 2{\sin ^2}y = \dfrac{1}{9} - 1
Take LCM on the right side,
⇒−2sin⁡2y=1−99\Rightarrow - 2{\sin ^2}y = \dfrac{{1 - 9}}{9}
Subtract the value in the numerator,
⇒−2sin⁡2y=−89\Rightarrow - 2{\sin ^2}y = - \dfrac{8}{9}
Now, divide both sides by -2,
⇒sin⁡2y=49\Rightarrow {\sin ^2}y = \dfrac{4}{9}
Take the square root on both sides,
⇒sin⁡y=49\Rightarrow \sin y = \sqrt {\dfrac{4}{9}}
As we know, the square root returns both positive and negative values. So,
⇒sin⁡y=±23\Rightarrow \sin y = \pm \dfrac{2}{3}
Substitute back the value of y,
⇒sin⁡(sin⁡−1x)=±23\Rightarrow \sin \left( {{{\sin }^{ - 1}}x} \right) = \pm \dfrac{2}{3}
We know that, sin⁡(sin⁡−1x)=x\sin \left( {{{\sin }^{ - 1}}x} \right) = x. Use this formula in the above equation,
∴x=±23\therefore x = \pm \dfrac{2}{3}
Thus, the value of x is ±23 \pm \dfrac{2}{3}.

Hence, option (C) is the correct answer.

Note: Whenever such a type of question appears, then always try to write either sine in terms of cosine or vice- versa, so that the whole equation converts into similar terms, which is easier to solve. In the solution, cosine is written in terms of sine.