If 2x2cos 60° – 4 cot245° – 2 tan60° = 0, what is the value of x?

Given, 2x2cos 60° – 4 cot245° – 2 tan60° = 0⇒2x2×12 - 412 - 23 = 0⇒x2 = 4 + 23⇒x2 = 3 + 12⇒x = 3 + 1

If 2x2cos 60° – 4 cot245° – 2 tan60° = 0, what is the value of x?

Given, 2x2cos 60° – 4 cot245° – 2 tan60° = 0⇒2x2×12 - 412 - 23 = 0⇒x2 = 4 + 23⇒x2 = 3 + 12⇒x = 3 + 1

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If 2x²cos 60° – 4 cot²45° – 2 tan60° = 0, what is the value of x?

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$1 - \sqrt{3}$

$3 - \sqrt{3}$

$\sqrt{3} + 1$

answer is D.

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Detailed Solution

Given, 2x²cos 60° – 4 cot²45° – 2 tan60° = 0

$\Rightarrow 2 x^{2} \times \frac{1}{2} - 4 {(1)}^{2} - 2 \sqrt{3} = 0$

$\Rightarrow x^{2} = 4 + 2 \sqrt{3}$

$\Rightarrow x^{2} = {(\sqrt{3} + 1)}^{2}$

$\Rightarrow x = \sqrt{3} + 1$

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If 2x2cos 60° – 4 cot245° – 2 tan60° = 0, what is the value of x?

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If 2x²cos 60° – 4 cot²45° – 2 tan60° = 0, what is the value of x?