How would I solve

$$a\cos^2(\theta) + b \sin^2(\theta) = 0$$

for $\theta$. Here $a,b$ are constants and $a \neq b, a \neq 0, b \neq0$.

I thought there might not be any solutions but with the constants $a,b,$ in front of them, I thought it is possible that they could be negative and the "right" constants to make the whole thing equal $0$.

So how would I solve this?

  • 1
    $\begingroup$ Use $\cos^2 + \sin^2 = 1$ $\endgroup$ – Stefan Mar 20 '13 at 14:59
  • $\begingroup$ Your guess is right @Kaish . In some cases there may not be any soln. $\endgroup$ – Abhra Abir Kundu Mar 20 '13 at 15:08

Hint: Write $\cos^2\theta=1-\sin^2\theta$.


Your equation is $b \sin^2 \theta=-a \cos^2 \theta$ or $\tan^2 \theta=-\frac{a}{b}$


$a\cos^2(\theta) + b \sin^2(\theta) = 0$

$a(1-\sin^2(\theta)) + b \sin^2(\theta) = 0$


$\sin \theta=\sqrt{\frac{a}{a-b}}$ or $\sin \theta=-\sqrt{\frac{a}{a-b}}$

There will be a soln. iff $-1\le |\sqrt{\frac{a}{a-b}}|\le 1$


You could use the double-angle formulas $\cos^2 \theta = \frac{1+\cos(2\theta)}{2}$, $\sin^2 \theta=\frac{1-\cos 2\theta}{2}$. Substituting these into your equation gives $$ \frac{a+b}{2}+\frac{a-b}{2} \cos (2\theta)=0 \, , $$ which can easily be solved for $\theta$.

Of course, this is equivalent to the solutions using the Pythagorean identity. But it has the advantage that you've traded in all your squares for doubled angles, so you don't have to worry about square roots in your solution...


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