Question

How do you solve `cos^2 x-sin^2 x=sin x`?

Answer 1

`(pi/6),(5pi)/6, (3pi)/2`

Explanation:

Replace in the equation `cos^2 x` by `(1 - sin^2 x)` -->
`1 - sin^2 x - sin^2 x - sin x = 0`
`-2sin^2 x - sin x + 1 = 0`
Solve this quadratic equation in sin x.
Since a - b + c = 0, use shortcut. The 2 real roots are: sin x = -1 and `sin x = -c/a = 1/2.`
a. `sin x = -1 --> x = 3pi/2`
b. `sin x = 1/2` --> 2 solution arcs -->
`x = pi/6`
`x = pi - pi/6 = (5pi)/6`
General answers:
`x = pi/6 + 2kpi`
`x = (5pi/6) + 2kpi`
`x = (3pi)/2 + 2kpi`
Checking these answers by calculator is advised.

Answer 2

`pi/6(4k+1),pi/2(4k-1)`

Explanation:

Putting `cos^2x-sin^2x=cos2x` we have

`cos^2x-sin^2x=sinx`

`=>cos2x=sinx`
`=>cos2x=cos(pi/2-x)`

So `2x=2kpi+-(pi/2-x)`

where `k epsilon Z`

when
`2x=2kpi+(pi/2-x)`
`=>2x+x=2kpi+pi/2=pi/2(4k+1)`
`=>3x=pi/2(4k+1)`
`=>x=pi/6(4k+1)`

Again when
`2x=2kpi-(pi/2-x)`
`=>2x=2kpi-pi/2+x`
`=>x=2kpi-pi/2=pi/2(4k-1)`