NIMCET 2020 Question Paper with solutions

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Question : 38 of 120

Marks: +1, -0

Solve the equation

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

for x on the interval

0≤x<2π

:

$\;{3 π}/{2}$
$\;{π}/{4}, \;{2 π}/{7}$
$\;{2 π}/{3}, \;{2 π}/{5}$
None of these

Solution:

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

Using the formula for the roots of a quadratic equation:

⇒ \sin x=\;{-(-1) ± √{(-1)^{2}-4(1)(-2)}}/{2(1)}

⇒ \sin x =\;{1+√{9}}/{2} OR \sin x =\;{1-√{9}}/{2}

⇒ \sin x =2 OR \sin x =-1

∵ -1 ≤ \sin x ≤ 1, ∴ \sin x =2

is not possible.

\sin x=-1=-\sin \;{π}/{2}=\sin (-\;{π}/{2})=\sin (2 n π-\;{π}/{2}) \;\; , n ∈ Z

For

n=0, x=-\;{π}/{2}

.
For

n =1, x =\;{3 π}/{2}

.
For

n =2, x =\;{7 π}/{2}

.
The only value of

x

on the interval

0 ≤ x <2 π

is

x =\;{3 π}/{2}

.

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