Finding the roots of a class of trignometric equations

Given,

$a \cos x + b \sin x + c \cos 2 x + d \sin 2 x + e = 0$

Letting $z = \tan \dfrac{x}{2}$ , and for simplicity, we'll assume that $x \in (-\pi , \pi )$ so that, $\dfrac{x}{2} \in (-\dfrac{\pi}{2}, \dfrac{\pi}{2} )$ , then,

$\cos x = \cos (2 \dfrac{x}{2} ) = 2 \cos^2 \dfrac{x}{2} - 1 = \dfrac{2}{1 + z^2} - 1 = \dfrac{1 - z^2}{1 + z^2}$

Moving on to $\sin x$ , we have that, $\sin^2 x = 1 - \cos^2 x = 1 - \dfrac{(1 - z^2)^2}{ (1 + z^2)^2} = \dfrac{4 z^2}{(1 + z^2)^2}$

So that,

$\sin x = \dfrac{2 z}{ 1 + z^2 }$

(Note from our definition of the range of $x$ , we have that $\sin x$ and $z$ have the same sign).

Next,

$\cos 2 x= 2 \cos^2 x - 1 = 2\dfrac{ (1-z^2)^2 }{(1+z^2)^2} - 1 = \dfrac{ 1 - 6 z^2 + z^4 } { (1 + z^2)^2 }$

And, finally,

$\sin 2 x = 2 \sin x \cos x = \dfrac{ 4 z (1 - z^2) }{ (1 + z^2)^2 }$

Using the above four expressions, our equation becomes, after multiplying through by $(1+z^2)^2$

$a (1 - z^2)(1+z^2) + b (2z)(1 + z^2) + c (1 - 6 z^2 + z^4) + d (4 z (1 - z^2) ) + e (1 + z^2)^2 = 0$

Simplifying,

$a (1 - z^4) + b ( 2 z + 2 z^3 ) + c (1 - 6 z^2 + z^4) + d ( 4 z - 4 z^3 ) + e (1 + 2 z^2 + z^4) = 0$

And finally, we have our quartic polynomial in $z$ ,

$z^4 ( -a + c + e ) + z^3 ( 2 b - 4 d) + z^2 ( -6 c + 2 e ) + z ( 2 b + 4 d ) + ( a + c + e ) = 0$

Plugging in the given values of $a, b, c, d, e$ , we find the real roots of this polynomial. For the given values, there are four real roots $z_i, \hspace{6pt} i = 1, 2, 3, 4$ , and for each of these four values, the corresponding value of the root $x_i$ is given by $x_i = 2 \tan^{-1} z_i$ . Now, since the required interval for the roots is $[0, 2 \pi )$ , then if $x_i \lt 0$ , we'll just add $2 \pi$ to it to make it lie in $[0, 2 \pi)$ .

Finally, we obtain, $\displaystyle \sum_{i=1}^4 x_i \approx 13.4936$

One thing to watch out for is the case when $x = \pi$ is a root of the equation, this can only happen if $-a + c + e = 0$ , but $(-a + c + e)$ is the coefficient of $z^4$ in the above quartic polynomial, so if this coefficient is zero, then we have to add $x = \pi$ to the real roots obtained from the resulting cubic polynomial.