How do you solve? #2(sin x)^2 − cos x − 1 = 0#

Question

Find all values of x in the interval [0, 2π] that satisfy the given equation. (Enter your answers as a comma-separated list.)

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Answer 1 · 2018-05-15T02:21:29

Given: #2sin^2(x) − cos(x) − 1 = 0#

Substitute: #sin^2(x) = 1-cos^2(x)#

#2(1-cos^2(x)) − cos(x) − 1 = 0#

#2-2cos^2(x) − cos(x) − 1 = 0#

#-2cos^2(x) − cos(x) +1 = 0#

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

Factor:

#(2cos(x)-1)(cos(x)+1)=0#

#cos(x) = 1/2 and cos(x) = -1#

Within the domain #[0, 2pi)#, the following values:

#x = pi/3,pi,(5pi)/3#

cause the original equation to be true.