I have this equation:

$$8\sin^2(x) + 6\cos^2(x) = 13\sin(2x)$$

I tried solving it, but the furthest I ever got was

$$4 - \cos^2(x) = 13\sin(x)\cos(x)$$

How can I solve this?

  • $\begingroup$ And yes, I tried using wolfram alpha, but the solution was 2 pages long, when this is a 10 minute problem according to my teacher. $\endgroup$
    – Bálint
    Commented Oct 26, 2017 at 9:56
  • 1
    $\begingroup$ One might observe that $4s^2-13sc+3c^2=(s-3c)(4s-c)$. $\endgroup$ Commented Oct 26, 2017 at 10:27

1 Answer 1


HINT: You can try dividing with $\cos^2(x)$ to get a quadratic in $\tan(x)$.


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