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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.

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  • $\begingroup$ Try to write it as $-8\sqrt{2}\sin^3\left(\dfrac\pi4-\dfrac x2\right)\sin\left(\dfrac x2\right)=0$, then it is much easier to solve. $\endgroup$ – Hakim Jul 20 '14 at 18:43
  • $\begingroup$ @Hakim That's what W|A gave me, but I couldn't understand how it got to that. $\endgroup$ – duci9y Jul 20 '14 at 18:45
  • $\begingroup$ I must admit that it isn't easy to arrive at that expression at first place, so you can use Vadim's strategy which is much simpler and requires much less work. $\endgroup$ – Hakim Jul 20 '14 at 18:46
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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 $\sin(\theta) + \cos(\theta) = 1$.

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  • $\begingroup$ Thanks a lot. Would this be the way to go for questions like this? $\endgroup$ – duci9y Jul 20 '14 at 18:55
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    $\begingroup$ You have two methods. One is to try and express everything in terms of $\sin(\theta)$ or everything in terms of $\cos(\theta)$. The other method would be similar to this method where you try to factor your expression. $\endgroup$ – Brad Jul 20 '14 at 18:57
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Strategy:

  1. Rewrite $\sin 2\theta=2\sin \theta \cos \theta$, and $\cos 2\theta=\cos^2\theta - \sin^2\theta$.

  2. Substitute $x=\cos\theta, y=\sin \theta$ to eliminate the trigonometry and get a conic section.

  3. Analyze the conic section to find $x,y$ that solve it. Hopefully not too many.

  4. Go back to $\theta$ by reversing the substitution from step 2.

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  • $\begingroup$ Uh oh… only if I had knowledge of conic sections… $\endgroup$ – duci9y Jul 20 '14 at 18:49
  • $\begingroup$ Luckily the quadratic factors; see @Brad's solution. $\endgroup$ – vadim123 Jul 20 '14 at 18:49

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