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I haven't practiced trigonometry for a while, and I'm wondering if there's a way to solve this analytically:

$$\sin(\alpha)-\cos(\alpha)\tan(\beta) = -a\frac{\tan(\beta)}{b}$$

I'm solving for $\alpha$, so $\beta$, $a$ and $b$ are known.

If there is, can someone point out the right way?

Thanks in advance!

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2 Answers 2

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Move the cosine term to the other side:$$\sin(\alpha)=\cos(\alpha)\tan(\beta) -a\frac{\tan(\beta)}{b}$$ Now square the equation (you will need to check if this introduce additional solutions) $$\sin^2(\alpha)=1-\cos^2\alpha=\left(\cos(\alpha)\tan(\beta) -a\frac{\tan(\beta)}{b}\right)^2$$ You can see now that you have a quadratic equation in $\cos\alpha$. Can you take it from here?

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  • $\begingroup$ I arrive to $$\frac{\frac{a}{b}tan^2 \beta \pm \sqrt(\tan^2 \beta (1-\frac{a^2}{b^2})+1)}{tan^2 \beta + 1}$$. The result for my problem isn't exactly what I was expecting but it is close enough. I'll take a closer look to see if I did something wrong, thanks! $\endgroup$ Commented Mar 24, 2021 at 15:13
  • $\begingroup$ Yes it works now. $\endgroup$ Commented Mar 24, 2021 at 15:47
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You want to solve $\sin(x)+u\cos(x)+v=0$. Writing $\cos(x)=y+y^{-1}$, $\sin(x)=\frac{y-y^{-1}}{i}$ with $y=e^{ix}$, giving you a degree two polynomial equation.$$ \frac{y-y^{-1}}{i} + u(y+y^{-1})+v=0 $$ So $$ y^2(u-i)+yv+(u+i)=0 $$ I think you can finish yourself.

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