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

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?

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?
• 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! Commented Mar 24, 2021 at 15:13
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.