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.