Solve sin^2 X+cos(2X)-cos X=0 over the domain 0º≤X<360º

Answer 1

sin^2(x) + cos(2x) -cosx= 0, 0</=x<2pi

cos2x = cos^2(x) - sin^2(x)

sin^2(x) +cos^2(x) -sin^2(x) - cosx = 0

cos^2(x) - cosx = 0

let y = cosx

y^2 - y = 0

factor

y(y-1) = 0

set each factor = 0

y=0, y-1=0

y = 0, 1= cosx

x = cos^-1(0), cos^-1(1) = pi/2, 0, 3pi/2 radians

or x= 90, 0, 270 degrees

Answer 2

sin²(x) + cos(2x) - cos(x) = 0

Using the Pythagorean identity sin²(x) + cos²(x) = 1, subtract cos²(x) from both sides to get:

sin²(x) = 1 - cos²(x) and substitute into the original equation:

[1 - cos²(x)] + cos(2x) - cos(x) = 0

Using the double angle identity cos(2x) = 2cos²(x) – 1, substitute into the above to get:

[1 - cos²(x)] + [2cos²(x) – 1] - cos(x) = 0

Simplify to get:

cos²(x) - cos(x) = 0

Factor out cos(x) to get:

cos(x)[cos(x) - 1] = 0

Set each piece equal to zero to get:

cos(x) = 0 and cos(x) - 1 = 0

To solve cos(x) = 0, consider the unit circle. cos(x) = 0 when x = 90° and 270°

To solve cos(x) - 1 = 0, add 1 to both sides then consider the unit circle. cos(x) = 1 when x = 0°

Solution: x = 0°, 90°, 270°