Question

How do you solve `Sin(theta)^2 - cos(theta)^2 = sin(theta)`?

Answer 1

`pi/2, (7pi)/6, (11pi)/6`

Explanation:

Use trig identity `cos 2a = cos^2 a - sin^2 a = 1 - 2sin^2` a, to transform the equation -->
`- (1 - 2sin^2 t) - sin t = 0`
`2sin^2 t - sin t - 1 = 0.`
Solve this quadratic equation for sin t
Since a + b + c = 0, use shortcut. the 2 real roots are: `sin t = 1` and `sin t = c/a = -1/2`.
Trig Table and unit circle -->
a. `sin t = 1` -> `t = pi/2`.
b. `sin t = -1/2` -> 2 solution arcs --> `t = - (pi/6)` and `t = (7pi)/6`
Since the arc `(11pi)/6` is co-terminal to arc `(-pi/6)`, therefor, the answers for `(0, 2pi)` are:
`pi/2, (7pi)/6, and (11pi)/6`

Answer 2

Alternative solution:

Explanation:

Consider the pythagorean identity `sin^2theta + cos^2theta = 1`. Rearrange this, solving for `-cos^2theta`, and you'll get: `-cos^2theta = sin^2theta - 1`. In order to change all terms to `sin` use this identity:

`sin^2theta + sin^2theta - 1 = sin theta`

Put all terms to one side of the equation.

`2sin^2theta - sin theta - 1 = 0`

Factor:

`2sin^2theta - 2sintheta + sin theta - 1 = 0`

`2sintheta(sin theta - 1) + 1(sin theta - 1) = 0`

`(2sintheta + 1)(sin theta - 1) = 0`

`sintheta = -1/2 and sin theta = 1`

By special angles:

`theta = 210˚ , 330˚ and 90˚`

Hopefully this helps!