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

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Answer 1 · 2016-04-05T03:33:26

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

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]}