I am solving a simple trig equation:

$$ 2\cos^2{\theta} + \cos{\theta} = 0 $$

My first approach was to move the $\cos\theta$ to the other side, and then divide by $\cos\theta$.

$$ 2\cos^2{\theta} = -\cos{\theta} $$ $$ 2\cos{\theta} = -1 $$ $$ \cos{\theta} = -\frac{1}{2} $$ $$ \theta = \frac{2\pi}{3}, \frac{4\pi}{3} $$

But apparently this approach causes me to miss some of the solution. The correct approach seems to be to extract $\cos\theta$ and solve like a quadratic:

$$ \cos{\theta}(2\cos{\theta}+1) = 0 $$ $$ \cos{\theta} = 0, -\frac{1}{2} $$ $$ \theta = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{2\pi}{3}, \frac{4\pi}{3} $$

I can kind of see that since $\cos\theta$ is squared, $\cos\theta$ could be $\pm$, but I'm not sure what it means for dividing by $\cos\theta$. Is it possible to come to the correct solution by dividing, or do I need to steer clear of that when it comes to squared values?

  • $\begingroup$ Note that when you factor and equate each to zero, you are checking both what you divided by as well as what you got after dividing. $\endgroup$ – dan post Apr 13 '18 at 1:58

The point is only that if $\cos\theta=0$ then you cannot divide both sides by $\cos\theta;$ i.e. you cannot divide both sides by $0.$ Note, for example that it is true that $3\times0=5\times0,$ but dividing both sides by $0$ and getting $3=5$ is wrong since $3\ne5.$

So dividing both sides by $\cos\theta$ gives you all solutions for which $\cos\theta\ne0,$ but then you also have to check whether there are any solutions for which $\cos\theta=0.$

If $AB=0,$ then either $A=0$ or $B=0.$ You have $A=\cos\theta$ and $B=2\cos\theta+1.$

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    $\begingroup$ ah awesome thank you! $\endgroup$ – Rob Allsopp Apr 13 '18 at 2:00

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