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

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

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