Approach:

The range of the trigonometric functions of \(\displaystyle{\sin{\theta}}\) is lie between \(\displaystyle{\left[-{1},{1}\right]}{\left[-{1},{1}\right]}\). No solution exists beyond this range.

Simplify the equation.

Obtain the factors of the equation.

The sum-to-product formulas for cosine is,

\(\displaystyle{\cos{{u}}}+{\cos{{v}}}=-{2}{\sin{{\frac{{{u}+{v}}}{{{2}}}}}}{\sin{{\frac{{{u}+{v}}}{{{2}}}}}}{\cos{{u}}}-{\cos{{v}}}=-{2}{\sin{{u}}}+{v}{2}{\sin{{u}}}-{v}{2}\)

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,

\(\displaystyle{\cos{{5}}}\theta-{\cos{{7}}}\theta={0}\)

\(\displaystyle-{2}{\sin{{\frac{{{5}\theta+{7}\theta}}{{{2}}}}}}{\sin{{\frac{{{5}\theta+{7}\theta}}{{{2}}}}}}={0}{\cos{{5}}}\theta-{\cos{{7}}}\theta={0}-{2}{\sin{{5}}}\theta+{7}\theta{2}{\sin{{5}}}\theta-{7}\theta{2}={0}-{2}{\sin{{6}}}\theta{\sin{{\left(-\theta\right)}}}={0}\)

\(\displaystyle-{2}{\sin{{6}}}\theta{\sin{{\left(-\theta\right)}}}={0}\)

Use the zero product property,

\(\displaystyle{\sin{{6}}}\theta={0}{\sin{{6}}}\theta={0}\ldots{\left({1}\right)}\ldots{\left({1}\right)}\)

\(\displaystyle{\sin{\theta}}={0}{\sin{\theta}}={0}\ldots{\left({2}\right)}\ldots{\left({2}\right)}\)

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,

\(\displaystyle{{\sin}^{{-{1}}}{\sin{\theta}}}={{\sin}^{{-{1}}}{\left({0}\right)}}\)

\(\displaystyle\theta={{\sin}^{{-{1}}}{\left({0}\right)}}{\sin{-}}{1}{\sin{\theta}}={\sin{-}}{1}{\left({0}\right)}\theta={\sin{-}}{1}{\left({0}\right)}\theta=\pi\)

\(\displaystyle\theta=\pi\)

The solution of the equation is obtained by adding in the integer multiples of \(\displaystyle\pi\pi\),

\(\displaystyle\theta={k}\pi\theta={k}\pi\)

Therefore, the solutions of the trigonometry equation \(\displaystyle{\cos{{5}}}\theta-{\cos{{7}}}\theta={0}{\cos{{5}}}\theta-{\cos{{7}}}\theta={0}\) is \(\displaystyle\theta=\pi{k},{\frac{{\pi{k}}}{{{6}}}}\theta=\pi{k},\pi{k}{6}\).