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.
Best 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 $\sin(\theta) + \cos(\theta) = 1$.