Solve the equation by first using a Sum-to-Product Formula. \cos 5 \theta - \cos 7 \theta = 0 \cos 5 \theta - \cos 7 \theta = 0

Using Sum-to-Product Formulas Solve the equation by first using a Sum-to-Product Formula.
$\mathrm{cos}5\theta -\mathrm{cos}7\theta =0\mathrm{cos}5\theta -\mathrm{cos}7\theta =0$
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Approach:
The range of the trigonometric functions of $\mathrm{sin}\theta$ is lie between $\left[-1,1\right]\left[-1,1\right]$. No solution exists beyond this range.
Simplify the equation.
Obtain the factors of the equation.
The sum-to-product formulas for cosine is,
$\mathrm{cos}u+\mathrm{cos}v=-2\mathrm{sin}\frac{u+v}{2}\mathrm{sin}\frac{u+v}{2}\mathrm{cos}u-\mathrm{cos}v=-2\mathrm{sin}u+v2\mathrm{sin}u-v2$
Cosine function has period $2\pi 2\pi$, thus find the solution in any interval of length $2\pi 2\pi$. Sine function is positive in first and second quadrant.
Calculation:
Consider the equation.
$\mathrm{cos}5\theta -\mathrm{cos}7\theta =0\mathrm{cos}5\theta -\mathrm{cos}7\theta =0$
Use Sum-to-Product formulas in the above equation,
$\mathrm{cos}5\theta -\mathrm{cos}7\theta =0$
$-2\mathrm{sin}\frac{5\theta +7\theta }{2}\mathrm{sin}\frac{5\theta +7\theta }{2}=0\mathrm{cos}5\theta -\mathrm{cos}7\theta =0-2\mathrm{sin}5\theta +7\theta 2\mathrm{sin}5\theta -7\theta 2=0-2\mathrm{sin}6\theta \mathrm{sin}\left(-\theta \right)=0$
$-2\mathrm{sin}6\theta \mathrm{sin}\left(-\theta \right)=0$
Use the zero product property,
$\mathrm{sin}6\theta =0\mathrm{sin}6\theta =0\dots \left(1\right)\dots \left(1\right)$
$\mathrm{sin}\theta =0\mathrm{sin}\theta =0\dots \left(2\right)\dots \left(2\right)$
Consider equation (1).
$\mathrm{sin}6\theta =0\mathrm{sin}6\theta =0$
Taking sine inverse both sides,
${\mathrm{sin}}^{-1}\mathrm{sin}6\theta ={\mathrm{sin}}^{-1}\left(0\right)$
$6\theta ={\mathrm{sin}}^{-1}\left(0\right)\mathrm{sin}-1\mathrm{sin}6\theta =\mathrm{sin}-1\left(0\right)6\theta =\mathrm{sin}-1\left(0\right)=0,\pi$
$=0,\pi$
The solution of the equation is obtained by adding in the integer multiples of $\pi$,
$6\theta =k\pi$
$\theta =\frac{k\pi }{6}6\theta =k\pi \theta =k\pi 6$
Consider equation (2).
$\mathrm{sin}\theta =0\mathrm{sin}\theta =0$
Taking $\mathrm{sin}$ inverse both sides,
${\mathrm{sin}}^{-1}\mathrm{sin}\theta ={\mathrm{sin}}^{-1}\left(0\right)$
$\theta ={\mathrm{sin}}^{-1}\left(0\right)\mathrm{sin}-1\mathrm{sin}\theta =\mathrm{sin}-1\left(0\right)\theta =\mathrm{sin}-1\left(0\right)\theta =\pi$