# 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# or#sin(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]}