How do you solve #cos 2x + sin x=0#?

2 Answers
Apr 14, 2015

#x=sin^-1(-1/2), x=sin^-1(1)#

Solution

#cos2x+sinx=0#

As

#cos2x=cos^2x-sin^2x#

So

#cos^2x-sin^2x+sinx=0#

#1-sin^2x-sin^2x+sinx=0#

#1-2sin^2x+sinx=0#

#-2sin^2x+sinx+1=0#

#2sin^2x-sinx-1=0#

#2sin^2x-2sinx+sinx-1=0#

#2sinx(sinx-1)+1(sinx-1)#

#(2sinx+1)(sinx-1)=0#

#(2sinx+1)=0, (sinx-1)=0#

#2sinx=-1, sinx=1#

#sinx=-1/2, sinx=1#

#x=sin^-1(-1/2), x=sin^-1(1)#

Apr 15, 2015

Replace #cos2x = 1 - 2sin^2 x#:

#f(x) = cos 2x + sin x = 1 - 2sin^2 x + sin x = 0#

Call #sin x = t#. This is a quadratic equation in #t#:

#f(t) = -2t^2 + t + 1 = 0#

Solve this quadratic equation. Because #a + b + c = 0#, one real root is #t_1 = 1# and the other is #t_2 = -1/2#

Next, solve the basic trig equation:

#t1 = sin x = 1 -> x = pi/2#

Solve:

#t2 = sin x = -1/2 -> x = (7pi)/6 and x = (11pi)/6#

Answers within period #(0, 2pi): pi/2; (7pi)/6; and (11pi)/6#

Extended answers:

#x = pi/2 + k*2pi#
#x = 7pi/6 + k*2pi#
#x = 11pi/6 + k*pi#
Check with f(x) = cos 2x + sin x = 0
#x= pi/2# --> #cos 2x = cos pi = -1#; #sin x = sin pi/2 = 1# --> f(x) = -1 + 1 = 0. OK
#x = (7pi)/6# --> #cos 2x = cos ((14pi)/6) = cos ((2pi)/6) = cos pi/3 = 1/2#; #sin ((7pi)/6) = -1/2# --> #f(x) = 1/2 - 1/2 = 0.# OK
#x = ((11pi)/6)# -> #cos ((22pi)/6) = cos ((10pi)/6) = cos ((5pi))/3 = 1/2#; #sin ((11pi)/6) = -1/2# --> #f(x) = 1/2 - 1/2 = 0 #. OK.