Solve for x:\(cos(2sin^{-1}x)=\frac{1}{9}\)

1.	Solve for x:\(cos(2sin^{-1}x)=\frac{1}{9}\)
Answer» To find: value of x Formula Used: \(2sin^{-1}x=sin^{-1}(2x\sqrt{1-x^2})\) Given: \(cos(2sin^{-1}x)=\frac{1}{9}\) LHS = cos(2sin^-1 x) Let θ = sin^-1x So, x = sin θ … (1) LHS = cos(2θ) = 1 – 2sin²θ Substituting in the given equation, \(1-2sin^2\theta=\frac{1}{9}\) \(2sin^2\theta=\frac{8}{9}\) \(sin^2\theta=\frac{4}{9}\) Substituting in (1), \(x^2=\frac{4}{9}\) \(x=\pm\frac{2}{3}\) Therefore,\(x=\pm\frac{2}{3}\) are the required values of x.

Answer»

To find: value of x

Formula Used: \(2sin^{-1}x=sin^{-1}(2x\sqrt{1-x^2})\)

Given: \(cos(2sin^{-1}x)=\frac{1}{9}\)

LHS = cos(2sin^-1 x)

Let θ = sin^-1x

So, x = sin θ … (1)

LHS = cos(2θ)

= 1 – 2sin²θ

Substituting in the given equation,

\(1-2sin^2\theta=\frac{1}{9}\)

\(2sin^2\theta=\frac{8}{9}\)

\(sin^2\theta=\frac{4}{9}\)

Substituting in (1),

\(x^2=\frac{4}{9}\)

\(x=\pm\frac{2}{3}\)

Therefore,\(x=\pm\frac{2}{3}\) are the required values of x.

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