Question

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

Answer 1

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

`(1-sin^2x)-sin^2x=sinx` `color(white)"ss"` (use Pythagorean Identity)

`1-2sin^2x=sinx`

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

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

`(2sinx-1)(sinx+1)=0` `color(white)"ss"` (it may help to substitute
`color(white)"ssssssssssssssssssssssssssssssssss"` `2S^2 + S -1 =0`
`color(white)"ssssssssssssssssssssssssssssssssss"` `(2S-1)( S +1) =0`

So `2sinx -1 =0` OR `sinx+1=0`
and `sinx = 1/2` OR `sinx = -1`

`sinx = 1/2` gives us `x= pi/6 + 2 pi k` or `(5 pi)/6+2 pi k`
`sinx = -1` gives us `x=(3 pi)/2 +2 pi k`
`color(white)"ssssssssssssssssssssssssss"` where `k` is any integer

Answer 2

Solve the equation: f(x) = cos^2 x - sin^2 x - sin x = 0.
Replace cos^2 x by (1 - sin^2 x)
f(x) = 1 - sin^2 x - sin^2 x - sin x = 0. Call t = sin x
Quadratic equation in t: f(t) = -2 t^2 - t + 1 = 0.
Solve this quadratic equation. There are 2 real roots : t1 = -1 and t2 = 1/2.
Solve the basic trig equation: t1 = sin x = -1 --> x = 3Pi/2
Solve t2 = sin x = 1/2 --> x = Pi/6 ; and x = 5Pi/6.
Within period (0. 2Pi), there are 3 answers: Pi/6; 5Pi/6; and 3Pi/2.