Question

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

Answer 1

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]}