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

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Answer 1 · 2018-04-15T22:29:26

I get $x \in {\frac{7 π}{12}, \frac{23 π}{12}}$ $x\in {{7\pi}/{12},{23\pi}/{12}}$ .

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

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

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

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

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

Possibility (1) gives

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

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

Possibility (2) gives

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

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