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$$3 \sin^2 \theta + 5 \sin \theta \cos \theta - 2\cos^2 \theta = 0$$

What are the steps to solve this equation for $ \theta $?

Because, I am always unable to deal with the product $\sin \theta \cos \theta$.

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  • $\begingroup$ Welcome to Math.SE! Please provide more context: what have you tried, where did you find this equation? $\endgroup$
    – Hrodelbert
    Commented Oct 15, 2015 at 9:38

6 Answers 6

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Try dividing by $\cos^2\theta$ all the equation...

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  • $\begingroup$ that would create fractions and then again, the same story. $\endgroup$
    – NubPro
    Commented Oct 15, 2015 at 10:08
  • $\begingroup$ No, everything would be in terms of the tangent function and can be solve as a quadratic equation. $\endgroup$
    – raul
    Commented Oct 15, 2015 at 10:10
  • $\begingroup$ oh snap. you're right! $\endgroup$
    – NubPro
    Commented Oct 15, 2015 at 10:21
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Use the linearisation formulae (inverse of the duplication formulae): \begin{align*} &=3\frac{1-\cos2\theta}2+\frac52\sin2\theta-2\frac{1+\cos2\theta}2\\ &=\frac12-\frac52(\cos2\theta-\sin2\theta)=\frac12-\frac{5\sqrt 2}2\cos\Bigl(2\theta-\frac\pi4\Bigr), \end{align*} whence \begin{align*} 3\sin^2\theta+&5\sin\theta\cos\theta-2\cos^2\theta=0\iff\cos\Bigl(2\theta-\frac\pi4\Bigr)=\frac1{5\sqrt2}\\ &\iff 2\theta-\frac\pi4\equiv\pm\arccos\frac1{5\sqrt2}\mod2\pi\\ &\iff \theta\equiv\frac\pi8\pm\frac12\arccos\frac1{5\sqrt2}\mod\pi. \end{align*}

A shorter method:

First observe we cannot have $\cos\theta=0$, for it would imply $\sin\theta=0$ and we cannot have both. So we can divide the equation by $ \cos\theta$, obtaining: $$3\tan^2\theta+5\tan\theta-2=0$$ Now the quadratic equation $t^2+5t-2=0$ has discriminant equal to $49$ and roots $\;\Bigl\{-2,\dfrac13\Bigr\}$. Thus we have to solve $$\begin{cases}\tan\theta=-2\\\tan\theta=\dfrac13\end{cases}\iff\begin{cases}\theta\equiv-\arctan2\mod\pi\\\theta\equiv\arctan\dfrac13\mod\pi=\dfrac\pi2-\arctan 3\mod\pi\end{cases}$$

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  • $\begingroup$ this is too superb for my standard. I do appreciate alternative methods. $\endgroup$
    – NubPro
    Commented Oct 15, 2015 at 10:02
  • $\begingroup$ @NubPro: I added another solution. Please see if if is more suitable. $\endgroup$
    – Bernard
    Commented Oct 15, 2015 at 10:18
  • $\begingroup$ Definitely. Thanks! $\endgroup$
    – NubPro
    Commented Oct 15, 2015 at 10:34
  • $\begingroup$ I think the short method here is the best way to tackle this problem down. $\endgroup$ Commented Oct 15, 2015 at 10:38
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Following excellent advice by @raul, you write immediately

$$\tan(\theta)=\frac{-5\pm\sqrt{5^2+4\cdot3\cdot2}}{2\cdot3}=\frac13,-2.$$

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  • $\begingroup$ yup...my bad... $\endgroup$
    – NubPro
    Commented Oct 15, 2015 at 10:27
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Here is a general strategy dealing with these kind of problems. Using summation formulas for sin and cos one can easily prove the following identities

$$\left\{ \matrix{ \cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta = 2{\cos ^2}\theta - 1 = 1 - 2{\sin ^2}\theta \hfill \cr \sin 2\theta = 2\sin \theta \cos \theta \hfill \cr} \right.$$

using these you can write

$$\eqalign{ & {\cos ^2}\theta = {{1 + \cos 2\theta } \over 2} \cr & {\sin ^2}\theta = {{1 - \cos 2\theta } \over 2} \cr & \sin \theta \cos \theta = {{\sin 2\theta } \over 2} \cr} $$

put these into your equation to get

$$3\left( {{{1 - \cos 2\theta } \over 2}} \right) + 5\left( {{{\sin 2\theta } \over 2}} \right) - 2\left( {{{1 + \cos 2\theta } \over 2}} \right) = 0$$

simplify to get

$$\eqalign{ & {1 \over 2} - {5 \over 2}\cos 2\theta + {5 \over 2}\sin 2\theta = 0 \cr & \sin 2\theta - \cos 2\theta + {1 \over 5} = 0 \cr} $$

the next step is to combine linear combinations of $\sin 2\theta $ and $\cos 2\theta $. You can do this again using summation formulas

$$\left\{ \matrix{ \sin x + w\cos x = \sin x + \tan \alpha \cos x = {1 \over {\cos \alpha }}\left( {\sin x\cos \alpha + \sin \alpha \cos x} \right) = {1 \over {\cos \alpha }}\sin \left( {x + \alpha } \right) \hfill \cr \tan \alpha = w \hfill \cr} \right.$$

In this case

$$\tan \alpha = - 1\,\,\, \to \,\,\,\alpha = {{3\pi } \over 4}\,\,\,\, \to \,\,\,\,\cos \alpha = - {1 \over {\sqrt 2 }}$$

and hence you can write

$$ - \sqrt 2 \sin (2\theta + {{3\pi } \over 4}) + {1 \over 5} = 0$$

or

$$\sin (2\theta + {{3\pi } \over 4}) = {1 \over {5\sqrt 2 }}$$

and hence

$$2\theta + {{3\pi } \over 4} = \left\{ \matrix{ \arcsin \left( {{1 \over {5\sqrt 2 }}} \right) \hfill \cr \pi - \arcsin \left( {{1 \over {5\sqrt 2 }}} \right) \hfill \cr} \right. + 2n\pi \,\,\,\,\,\,\,\,\,\,\,\,\,n = 1,2,3,...$$

which finally gives

$$\theta = {1 \over 2}\left( {\left\{ \matrix{ \arcsin \left( {{1 \over {5\sqrt 2 }}} \right) \hfill \cr \pi - \arcsin \left( {{1 \over {5\sqrt 2 }}} \right) \hfill \cr} \right. + 2n\pi - {{3\pi } \over 4}} \right)$$

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  • $\begingroup$ Well, unfortunately my studies doesn't cover thoroughly at this point. $\endgroup$
    – NubPro
    Commented Oct 15, 2015 at 10:20
  • $\begingroup$ you mean you don't know summation formulas for sin and cos? The base of all I have used are these formulas. :) $\endgroup$ Commented Oct 15, 2015 at 10:23
  • $\begingroup$ arcsin thingy.. $\endgroup$
    – NubPro
    Commented Oct 15, 2015 at 10:25
  • $\begingroup$ As an example, how do you solve $sin(x)=0.1$? $\endgroup$ Commented Oct 15, 2015 at 10:28
  • $\begingroup$ omg.....I figured out what arcsin is. XD This is embarrassing. Do you pronounce it as "ARC SIN" or "INVERSE SIN"? $\endgroup$
    – NubPro
    Commented Oct 15, 2015 at 10:32
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Factorise this into the quadratic:

$$3 \sin^2 \theta + 5 \sin \theta \cos \theta - 2\cos^2 \theta = 0$$

$$(3 \sin \theta - \cos \theta )(sin \theta + 2\cos \theta) = 0$$

This gives 2 equations.

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

You should be able to solve it by dividing both by cos. And then solve for tan.

(Make sure you check the cases if $$\cos \theta = 0$$ separately afterwards to see if you missed any solutions when dividing by cos)

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  • $\begingroup$ My apologies, got the signs wrong on the 1st answer. Edited it. $\endgroup$
    – Shuri2060
    Commented Oct 15, 2015 at 9:47
  • $\begingroup$ What do you mean by checking the cases? $\endgroup$
    – NubPro
    Commented Oct 15, 2015 at 10:00
  • $\begingroup$ Typo in your fourth equation. $\endgroup$
    – user65203
    Commented Oct 15, 2015 at 10:21
  • $\begingroup$ Dividing by cos means that you will miss any solutions when cos theta = 0. To avoid this, just simply see what happens if cos theta = 0 in the original equation. If cos theta = 0 works to solve the equation, then it is a solution. If not, then you haven't missed any extra solutions (Note that in this case, cos theta = 0 means that sin theta = 1, in which case you won't get any extra solutions) $\endgroup$
    – Shuri2060
    Commented Oct 15, 2015 at 14:05
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$3\sin^2{\theta}-2\cos^2{\theta}+5\sin{\theta}\cos{\theta}=0$
Transfer the $\sin{\theta}\cos{\theta}$ term to the RHS.
$3\sin^2{\theta}-2\cos^2{\theta}=-5\sin{\theta}\cos{\theta} $
Then square both sides.
$(3\sin^2{\theta}-2\cos^2{\theta})^2=(-5\sin{\theta}\cos{\theta})^2$
$9\sin^4{\theta}-12\sin^2{\theta}\cos^2{\theta}+4\cos^4{\theta}=25\sin^2{\theta}\cos^2{\theta}$
Then convert all occurrences of $\sin^2{\theta}$ to $1-\cos^2{\theta}$.
$9(1-\cos^2{\theta})^2-12(1-\cos^2{\theta})\cos^2{\theta}+4\cos^4{\theta}=(1-\cos^2{\theta})\cos^2{\theta}$
This will give an equation which can be solved for $\cos^2{\theta}$,
$50\cos^4{\theta}-55\cos^2{\theta}+9=0$
then for $\cos{\theta}$,
then for $\theta$.

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