How do you solve cos2x + sinx = 0 over the interval 0 to 2pi?
1 Answer
Change
Explanation:
Double angle formula for cosine
cos(2x) -= 1 - 2 sin^2(x)
Replace the
cos(2x) + sin(x) = 1 - 2 sin^2(x) + sin(x)
= 0
Solve the quadratic by factorizing
-2 sin^2(x) + sin(x) + 1 = 0
(1 + 2sin(x))(1 - sin(x)) = 0
Therefore, either
sin(x) = 1 orsin(x) = -1/2
For
x = pi + pi/6 = (7pi)/6 or
x = 2pi - pi/6 = (13pi)/6
Putting together all the answers,
Refer to the
graph{cos(2x)+sin(x) [-10, 10, -5, 5]}