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

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

1 Answer
May 15, 2018

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.