Question

+.UT Find all real numbers in the interval [0.21) that satisfy the equation. Use radian measure. sin x cos WI + cos x sin V2 2 3 The solution set is O. (Simplify your answer. Type an exact answer, using a as needed. Use integers or fractions for any numbers in

Answer 1

$\sin(x+y)=\sin x\cos y+\cos x\sin y$

$\text{LHS, Given }\sin x\cos\left ( \frac{\pi}{3} \right )+\cos x\sin\left ( \frac{\pi}{3} \right )\\ \text{ Satsifies the properties of the above identity. Hence,}\\ \sin x\cos\left ( \frac{\pi}{3} \right )+\cos x\sin\left ( \frac{\pi}{3} \right )=\sin\left (x+\frac{\pi}{3} \right )$

Now,

$\bigskip \sin\left (x+\frac{\pi}{3} \right )=\frac{\sqrt2}{2} \\\bigskip\Rightarrow \sin\left ( x+\frac{\pi}{3} \right )=\frac{1}{\sqrt 2} \\\bigskip\Rightarrow \sin\left ( x+\frac{\pi}{3} \right )=\sin\left ( \frac{\pi}{4} \right ) \\\bigskip\Rightarrow \left ( x+\frac{\pi}{3} \right )=n\pi +(-1)^n\frac{\pi}{4}\enspace \text{ where }n\in\mathbb{Z} \\\bigskip\Rightarrow x=n\pi -\frac{\pi}{3}+(-1)^n\frac{\pi}{4}\enspace\text{ where }n\in\mathbb{Z}$

Thus, the solution set is

$\text{Since }x\in[0,2\pi),\text{ we obtain the solution set of x by putting }n=1,\enspace 2$

Therefore, the solution set of x is

${\color{Blue} \left \{ \frac{5\pi}{12}, \frac{23\pi}{12} \right \}}$