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.