Solve the equation ²x- x-2=0 for x on the interval 0≤ x

Correct Answer is : 3π23π/223π

Correct Answer is : 3π23π/223π

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NIMCET 2020 Question Paper with solutions

Question : 38 of 120

Marks: +1, -0

\sin^2x-\sin x-2=0

0\leq x<2\pi

Solution:

\sin^{2}x-\sin x-2=0

Using the formula for the roots of a quadratic equation:

\Rightarrow \sin x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-2)}}{2(1)}

\Rightarrow \sin x = \frac{1+\sqrt{9}}{2} \text{ OR } \sin x = \frac{1-\sqrt{9}}{2}

\Rightarrow \sin x = 2 \text{ OR } \sin x = -1

\because -1 \le \sin x \le 1, \therefore \sin x =2

is not possible.

\sin x = -1 = -\sin \frac{\pi}{2} = \sin\left(-\frac{\pi}{2}\right) = \sin\left(2n\pi - \frac{\pi}{2}\right)\;,\; n \in \mathbb{Z}

For

n=0, x=-\frac{\pi}{2}

For

n=1, x=\frac{3\pi}{2}

For

n=2, x=\frac{7\pi}{2}

The only value of

x

on the interval

0 \leq x < 2\pi

x = \frac{3\pi}{2}

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