How do you solve cos2x + sinx = 0 over the interval 0 to 2pi?

1 Answer
Apr 5, 2016

Change cos(2x) to sin(x) using the double angle formula and solve the resulting quadratic equation.

Explanation:

Double angle formula for cosine

cos(2x) -= 1 - 2 sin^2(x)

Replace the cos(2x) in the equation.

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 or sin(x) = -1/2

sin(x) = 1 corresponds to x = sin^{-1}(1) = pi/2.

For sin(x) = -1/2, the basic angle is sin^{-1}(1/2) = pi/6. Since sin(x) is negative when x is in the third or forth quadrant,

x = pi + pi/6 = (7pi)/6

or

x = 2pi - pi/6 = (13pi)/6

Putting together all the answers, x = pi/2, (7pi)/6 or (13pi)/6.

Refer to the x-intercepts of the graph y = cos(2x) + sin(x) below.
graph{cos(2x)+sin(x) [-10, 10, -5, 5]}