Question

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

Answer 1

`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)`

Answer 2

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.