JEE MAIN - Mathematics (2021 - 31st August Morning Shift - No. 9)
cosec18$$^\circ$$ is a root of the equation :
x2 + 2x $$-$$ 4 = 0
4x2 + 2x $$-$$ 1 = 0
x2 $$-$$ 2x + 4 = 0
x2 $$-$$ 2x $$-$$ 4 = 0
Explanation
$$\cos ec18^\circ = {1 \over {\sin 18^\circ }} = {4 \over {\sqrt 5 - 1}} = \sqrt 5 + 1$$
Let $$\cos ec18^\circ = x = \sqrt 5 + 1$$
$$ \Rightarrow x - 1 = \sqrt 5 $$
Squaring both sides, we get
$${x^2} - 2x + 1 = 5$$
$$ \Rightarrow {x^2} - 2x - 4 = 0$$
Let $$\cos ec18^\circ = x = \sqrt 5 + 1$$
$$ \Rightarrow x - 1 = \sqrt 5 $$
Squaring both sides, we get
$${x^2} - 2x + 1 = 5$$
$$ \Rightarrow {x^2} - 2x - 4 = 0$$
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