How do you find all values of x in the interval [0, 2pi] in the equation #2sin^2x - sinx - 1 = 0#?

1 Answer
Shwetank Mauria
Mar 27, 2016

Solution set in the interval #[0,2pi]# is #{pi/2,(7pi)/6,(11pi)/6}#

Explanation:

To find all values of x in the interval #[0, 2pi]# in the equation #2sin^2x−sinx−1=0#, let us solve by considering #sinx# as variable and using quadratic formula.

Hence #sinx=(-(-1)+-sqrt((-1)^2-4xx2xx(-1)))/(2xx2)# or

#sinx=(1+-sqrt(1+8))/4# i.e.

#sinx=(1+-3)/4# i.e. #sinx=1# or #-1/2#

In the interval #[0,2pi]#, #sinx=1# only for #x=pi/2#

and #sinx=-1/2# for #x=(7pi)/6# and #(11pi)/6#

Hence solution set in the interval #[0,2pi]# is #{pi/2,(7pi)/6,(11pi)/6}#

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