Answer :

Put x = cos2θ

We know that cos2θ = 1 – 2sin^{2}θ and cos2θ = 2cos^{2}θ – 1

Hence 1 + cos2θ = 2cos^{2}θ and 1 – cos2θ = 2sin^{2}θ

⇒ 1 + x = 1 + cos2θ = 2cos^{2}θ

⇒ 1 – x = 1 – cos2θ = 2sin^{2}θ

Consider LHS

Cancel out √2

We know that

Using

As x = cos2θ ⇒ cos^{-1}x = 2θ ⇒ θ = 1/2(cos^{-1}x)

⇒ LHS = RHS

**Hence proved**

**OR**

We know that

Here and

As

⇒ x^{2} + 4x – 2x – 8 + x^{2} – 4x + 2x – 8 = x^{2} – 16 – (x^{2} – 4)

⇒ 2x^{2} – 16 = x^{2} – 16 – x^{2} + 4

⇒ 2x^{2} = 4

⇒ x^{2} = 2

⇒ x = ±√2

**Hence x is ±√2.**

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