trigonometry - How to solve $3 - 2 cos theta - 4 sin theta - cos 2theta + sin 2theta = 0$

I have got a bunch of trig equations to solve for tomorrow, and got stuck on this one.

Solve for $\theta$:

$$3 - 2 \cos \theta - 4 \sin \theta - \cos 2\theta + \sin 2\theta = 0$$

I tried using the addition formula, product-to-sum formula, double angle formula and just brute force by expanding all terms on this, but couldn't get it.

I am not supposed to use inverse functions or a calculator to solve this.

Tried using Wolfram|Alpha's step by step function on this, but it couldn't explain things.

Answer

Let $x = \sin(\theta), y = \cos(\theta)$

$$3 - 2 y - 4x - 2y^2+1 + 2xy = 0$$

Simplify, divide by $2$ and replace $y^2$ with $1-x^2$.

$$1 - y - 2x+x^2+ xy = 0$$

Factor

$$(x-1)(x+y-1) = 0$$

Now just solve $\sin(\theta) = 1$ and $\sin(\theta) + \cos(\theta) = 1$.