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In the equation $x\cos(\theta) + y\sin(\theta) = z,$ how do I solve in terms of $\theta$? i.e $\theta = \dots$.

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    Linear equations in $\sin \theta$ and $\cos \theta$ can be solved by a resolvent quadratic equation. One method is to write the $\sin \theta$ and $\cos \theta$ functions in terms of the $\tan (\theta/2)$.2011-04-15

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There are various possible strategies. I will mention one approach. Of course if there is a solution, there are infinitely many, since we can add $2\pi$ to any solution and get another solution.

Let's change notation a little. We are interested in the equation $a\cos\theta + b\sin\theta=q$

Rewrite this equation as $\frac{a}{\sqrt{a^2+b^2}}\cos\theta+ \frac{b}{\sqrt{a^2+b^2}}\sin\theta=\frac{q}{\sqrt{a^2+b^2}}$ Now let $\phi$ be the angle whose sine is $a/\sqrt{a^2+b^2}$ and whose cosine is $b/\sqrt{a^2+b^2}$. By a formula that I hope is familiar (sine of a sum of angles), the equation can be rewritten as $\sin(\phi+\theta)=\frac{q}{\sqrt{a^2+b^2}}$ Look at the right-hand side. If its absolute value is greater than $1$, there will be no (real) solution. Otherwise, for simplicity, call the right-hand side $w$. Then we can write that $\phi+\theta=\arcsin w$ or $\phi+\theta=\pi-\arcsin w$. Now remember that whatever solutions you get through this process, anything obtained by adding $2n\pi$, where $n$ is an integer, to a solution, is also a solution.

The reason I went in detail through this approach is that in Physics, it is often important to express $a\cos\theta+b\sin\theta$ in the form $c\sin(\phi+\theta)$ that we used to solve the equation. There are other approaches.

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    @Didier Pau: $S$orry, I missed part of the content of your comment. I have edited the answer to take account of what you wrote.2011-04-15
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This expands my comment above. As I wrote here "certain trigonometric equations such as the linear equations in $\sin x$ and $\cos x$ can be solved by a resolvent quadratic equation. One method is to write the $\sin x$ and $\cos x$ functions in terms of (...) $\tan$ of the half-angle".

Applying this method, since

$\cos \theta =\frac{1-\tan ^{2}\frac{\theta }{2}}{1+\tan ^{2}\frac{\theta }{2% }}\qquad\text{and}\qquad\sin \theta =\frac{2\tan \frac{\theta }{2}}{1+\tan ^{2}\frac{\theta }{2}},$

your equation is equivalent to

$x\frac{1-\tan ^{2}\frac{\theta }{2}}{1+\tan ^{2}\frac{\theta }{2}}+y\frac{% 2\tan \frac{\theta }{2}}{1+\tan ^{2}\frac{\theta }{2}}=z.$

Let $u=\tan \frac{\theta }{2}$. Then we can write it as

$\left( x+z\right) u^{2}-2yu+z-x=0,$

which has the solutions

$u=\tan \frac{\theta }{2}=\frac{1}{2\left( x+z\right) }\left( 2y\pm2\sqrt{% y^{2}+x^{2}-z^{2}}\right).$

Thus

$\theta =2\arctan \left( \frac{1}{ x+z }\left( y\pm % \sqrt{y^{2}+x^{2}-z^{2}}\right) \right) +2n\pi,\qquad n\in\mathbb{Z} .$

This method is valid iff $\theta \neq (2k+1)\pi $, with $k\in\mathbb{Z}$.

A different technique is to use an auxiliary angle.

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    You are welcome! I adapted this technique from my 1967 Trigonometry text book (J. Calado, Compêndio de Trigonometria), where it is presented and worked out in detail. I deleted my explanation of the auxiliary angle technique (answered before by user6312).2011-04-15
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Let us introduce $c=\cos \theta$. Then your equation reads $x c + s y \sqrt{1-c^2} =z$ where $s=\pm 1$ is related to the quadrant of $\theta$ (+1 in the first and second quadrant and -1 in the third and forth). Subtracting $x c$ from both sides then squaring the equation yields $y^2 (1-c^2) = (z- xc)^2 = z^2 -2 zx c + x^2 c^2.$ This is a quadratic equation with the solutions $ c_\pm = \frac{x z \pm y \sqrt{x^2+ y^2 - z^2}}{x^2 + y^2}.$ In order that $c_\pm$ are real, we need $z^2 \leq x^2+y^2$. As we have squared the equation, we have to check whether $c_\pm$ solves the original equation. Indeed, $c_\pm$ solves the original equation with $s_\pm=\text{sgn}[(z-xc_\pm)/y]$. Therefore, we have the two solutions (mod $2\pi$) $\theta = s_\pm \arccos(c_\pm). $

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    The signs one has to avoid to get values which can be solutions are easy to determine (this is your last comment). But the problem is to be sure that the remaining ones actually yield solutions (this was my remark). (In case you are wondering, I know how to solve this exercise, thank you. My point is, too many steps in your post are left vague for it to be convincing.)2011-04-15