Find all values of x in the interval [0, 2π] that satisfy the given equation? 2(sin x)^2 − cos x − 1 = 0

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Answer 1 · 2018-04-12T12:28:03

#color(blue)(pi/3,pi,(5pi)/3)#

Identity:

#color(red)bb(sin^2x+cos^2x=1)#

#2sin^2x-cosx-1=0#

From the identity:

#sin^2x=1-cos^2x#

#2(1-cos^2x)-cosx-1=0#

Expanding bracket:

#2-2cos^2x-cosx-1=0#

Simplifying:

#2cos^2x+cosx-1=0#

Let #u=cosx#

#:.#

#2u^2+u-1=0#

Factor:

#(2x-1)(x+1)=0=>u=1/2 and u=-1#

#:.#

#cosx=1/2 and cosx=-1#

#x=arccos(cosx)=arccos(1/2)=>x=pi/3,(5pi)/3#

#x=arccos(cosx)=arccos(-1)=>x=pi#

Solution are therefore:

#color(blue)(pi/3,pi,(5pi)/3)#