How do you solve cos 2x + sin x=0?
2 Answers
Solution
As
So
Replace
f(x) = cos 2x + sin x = 1 - 2sin^2 x + sin x = 0
Call
f(t) = -2t^2 + t + 1 = 0
Solve this quadratic equation. Because
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
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.