Approach:

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

Simplify the equation.

Obtain the factors of the equation.

Use sine Double-Angled formulas,

\(\displaystyle{\sin{{2}}}{x}={2}{\sin{{x}}}{\cos{{x}}}\)

Cosine and sine functions has period of \(\displaystyle{2}\pi\), thus find the solution in any interval of length \(\displaystyle{2}\pi\)

Cosine function is positive in first and fourth quadrant. SIne function is positive in first and second quadrant.

Calculation:

Consider the equation.

\(\displaystyle{\sin{\theta}}-{\cos{\theta}}={\frac{{{1}}}{{{2}}}}\)

Squaring both sides in above equation,

\(\displaystyle{\left({\sin{{t}}}{h}{e}{t}-{\cos{\theta}}\right)}^{{{2}}}={\left({\frac{{{1}}}{{{2}}}}\right)}^{{{2}}}\)

\(\displaystyle{{\cos}^{{{2}}}\theta}+{{\sin}^{{{2}}}\theta}-{2}{\cos{\theta}}{\sin{\theta}}={\frac{{{1}}}{{{4}}}}\)

\(\displaystyle{1}-{2}{\cos{\theta}}{\sin{\theta}}={\frac{{{1}}}{{{4}}}}\)

Use doubled-Angled the equation,

\(\displaystyle{\sin{{2}}}\theta={\frac{{{3}}}{{{4}}}}\)

Taking sine inverse both sides,

\(\displaystyle{{\sin}^{{-{1}}}{\sin{{2}}}}\theta={{\sin}^{{-{1}}}{\left({\frac{{{3}}}{{{4}}}}\right)}}\)

\(\displaystyle{2}\theta={{\sin}^{{-{1}}}{\left({\frac{{{3}}}{{{4}}}}\right)}}\)

\(\displaystyle{2}\theta\approx{0.85}\)

\(\displaystyle\theta\approx{0.43}\)

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

\(\displaystyle\theta\approx{0.43}+{k}\pi\)

Substitute \(\displaystyle{k}={0},{1},{2}\)

\(\displaystyle\theta\approx{0.43}+{0}={0.43}\)

\(\displaystyle\theta\approx{0.43}+\pi={3.56}\)

\(\displaystyle\theta\approx{0.43}+{2}\pi={1.15}\)

From the above obtained solutions only 1.15 and is 3.56 satisfying the equation, so these are the only solutions of the equation.

Therefore, the solution of the trigonometry equation \(\displaystyle{\sin{\theta}}-{\cos{\theta}}={\frac{{{1}}}{{{2}}}}\) in the interval \(\displaystyle{\left[{0},{2}\pi\right)}\) is \(\displaystyle\theta={1.15},{3.56}\).

Conclusion:

Hence, the solution of the trigonometry equation \(\displaystyle{\sin{\theta}}-{\cos{\theta}}={\frac{{{1}}}{{{2}}}}\) in the interval \(\displaystyle{\left[{0},{2}\pi\right)}\) is \( \theta=1.15, 3.56\).