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Consider the equation $$ \cos^2\phi + \alpha\sin\phi\cos\phi-\beta=0\;, $$ where $\alpha,\beta\in\mathbb{R}$. I need to find an explicit expression for $\phi$. I have tried completing the square, but that did not go far. Any ideas are welcome.

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6 Answers 6

Use the two trigonometric identities $$ \cos^2\varphi = \frac 1 2 + \frac 1 2 \cos(2\varphi)\qquad \text{ and } \qquad 2\sin\varphi\cos\varphi = \sin(2\varphi). $$ This transforms your equation to $$ \frac 1 2 + \frac 1 2 \cos(2\varphi) + \frac 1 2 \alpha\sin(2\varphi) - \beta = 0, $$ or $$ \cos(2\varphi)+ \alpha\sin(2\varphi) = 2\beta - 1. $$ Now you have a linear combination of sine and cosine functions that both have the same period. The coefficients are $1$ and $\alpha$, and you want a linear combination in which the sum of square of the coefficients is $1$, so that one of them will be a sine and the other a cosine. So you can write $$ \sqrt{1+\alpha^2}\left( \frac{1}{\sqrt{1+\alpha^2}}\cos(2\alpha)+ \frac{\alpha}{\sqrt{1+\alpha^2}} \sin(2\alpha) \right) = 2\beta - 1. $$ This becomes $$ \sqrt{1+\alpha^2}\Big( \sin(\chi)\cos(2\alpha)+ \cos(\chi) \sin(2\alpha) \Big) = 2\beta - 1 $$ and then $$ \sqrt{1+\alpha^2} \sin(2\varphi+\chi) = 2\beta - 1. $$

Can you take it from there?

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Multiply by $2$. Then note that $2 \cos ^2 x = \cos 2x +1$. Let $\cos 2 \phi = y$. Then we have $y+\alpha \sqrt{1-y^2}+c = 0 ,(c = 1- \beta)$ Wolfram Alpha solves this as $\cos 2 \phi = \frac{c \pm \sqrt{\alpha4- \alpha^2 c^2+\alpha^2}}{\alpha^2+1}$.

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My answer is a little tedious, which involves the discussion of the values of $\alpha$ and $\beta$.

First, you can use trigonometric identities to change your equation into: $$ \frac{1+ \cos{2 \phi}}{2} +\frac{\alpha}{2} \sin{2 \phi} - \beta = 0$$ Moving $\cos{2\phi}$ to the right hand side and taking squares of both side gives you $$(1+\alpha^2)\cos^2{2\phi} - 2(2\beta -1) \cos{2\phi} + (2\beta-1)^2 - \alpha^2=0$$ which can be solved as a quadratic function. Note that you need to discuss the existence of solution, since $|\cos{2\phi}| \leq 1$. Once you get the solution to $\cos{2\phi}$, it is easy to get the expression of $\phi$.

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Divide either sides by $\cos^2\phi$ to get

$$1+a\tan\phi=\frac b{\cos^2\phi}=b\sec^2\phi=b(1+\tan^2\phi)$$

$$\iff b\tan^2\phi-a\tan\phi+b-1=0$$ which is a Quadratic Equation in $\tan\phi$

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Hint: $\cos^2 \phi=\dfrac{\cos 2\phi +1}2$ and $\sin \phi \cos\phi=\dfrac{\sin 2\phi}2$.

Then your expression is a linear function of $\cos 2\phi$ and $\sin 2\phi$.

Do you know how to solve $a\sin x +b\cos x=c$ for $x$?

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Another way is to use the trigonometric identity:- $1 = sin^2\phi + \cos^2\phi$.

Then, the original equation can be re-written as $\cos^2\phi + \alpha\sin\phi\cos\phi-\beta (sin^2\phi + \cos^2\phi)=0$

After re-arranging terms, we have $(1 - \beta)\cos^2\phi + \alpha\sin\phi\cos\phi-\beta sin^2\phi=0$

Which is a (homogeneous) quadratic equation solvable by (1) applying the quadratic formula appropriately or (2) factorization (if $\alpha$ and $\beta$ are 'well behave' known numerical quantities.

The identity:- $\frac {sin \theta} {cos \theta} = tan \theta$ is also needed to get the value(s) of $\theta$.

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