# Find all real numbers in interval [0,2π)? sin x cos(π/4)+cos x sin (π/4)=1/2

Apr 15, 2018

I get $x \setminus \in \setminus \left\{\frac{7 \setminus \pi}{12} , \frac{23 \setminus \pi}{12} \setminus\right\}$.

#### Explanation:

The expression given fits the formula for the sine of a sum:

$\setminus \sin \left(x + y\right) = \setminus \sin \left(x\right) \setminus \cos \left(y\right) \setminus + \cos \left(x\right) \setminus \sin \left(y\right)$

Here put in $y = \frac{\setminus \pi}{4}$ and $\setminus \sin \left(x + y\right) = \frac{1}{2}$. Then we have the following possibilities:

(1) $x + \frac{\setminus \pi}{4} = \frac{\setminus \pi}{6} + 2 m \setminus \pi$

(2) $x + \frac{\setminus \pi}{4} = \frac{5 \setminus \pi}{6} + 2 n \setminus \pi$

Possibility (1) gives

$x = - \frac{\setminus \pi}{12} + 2 m \setminus \pi$

and to get a number between $0$ and $2 \setminus \pi$ we take $m = 1$, thus $x = \frac{23 \setminus \pi}{12}$.

Possibility (2) gives

$x = \frac{7 \setminus \pi}{12} + 2 n \setminus \pi$

and to get a number between $0$ and $2 \setminus \pi$ we take $n = 0$, thus $x = \frac{7 \setminus \pi}{12}$.