How do you solve the following equation #2 sin ^2 x - sin x - 1 = 0# in the interval [0, 2pi]?

1 Answer
Alan P. · Zor Shekhtman
Jun 18, 2016

#x in {pi/2, (7pi)/2, (11pi)/2}#

Explanation:

Given #2sin^2(x)-sin(x)-1=0#

Replacing #sin(x)# with #h#
#color(white)("XXX")2h^2-h-1=0#
then using the quadratic formula for roots
#color(white)("XXX")h=(1+-sqrt((-1)^2-4(2)(-1)))/(2(2))#

#color(white)("XXXX")=(1+-sqrt(9))/4#

#color(white)("XXXX")=1" or " -1/2#

within the interval #[0,2pi]#
#{: ("if "sin(x)=1,color(white)("XXXXXX"),"if "sin(x)=-1/2), (color(white)("X")x=pi/2,,color(white)("X")x=(7pi)/6" or " (11pi)/6) :}#
(Note that these are standard angles for which you should know or be able to figure out the trigonometric ratios).

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