Find the general solution of the equation 2sin2(x)−cos(x)−1=0

general-solution
math2sin2x-cosx-10

#1

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Answer:

From the identity sin^2(x) + cos^2(x) = 1, you get sin^2(x) = 1 - cos^2(x). Plug in into the equation above, you get:
2(1 - cos^2(x)) - cos(x) - 1 = 0
2 - 2cos^2(x) - cos(x) - 1 = 0
-2cos^2(x) - cos(x) + 1 = 0
Multiply throughout by -1,
2cos^2(x) + cos(x) - 1 = 0

Put y = cos(x), so you get
2y^2 + y - 1 = 0

Factor the quadratic equation:
2y^2 + 2y - y - 1 = 0
2y(y + 1) - 1(y + 1) = 0
(2y - 1)(y + 1) = 0
y = 1/2 and y = -1

Again put cos(x) = y,
cos(x) = 1/2 and cos(x) = -1

cos(x) equals 1/2 when x equals 60 degree angle or pi/3 so x = pi/3

cos(x) equals -1 when x equals pi so x = pi

Whole both of you are nearly correct, the “general solution” should include every solution.
So, every event that yields cos(x) = 1/2 ,1.
x=+/- π/3 and 1, and everywhere it repeats.
x= +/-π/3(k) +2π and x= π(k) + π for all integer value of k.