Solve-it-in-pi-pi-sin-2x-cos-x-

Question Number 75502 by mathocean1 last updated on 12/Dec/19

S olve it in] - π; π]

\sin (2 x) = \cos (x)

Commented by mathmax by abdo last updated on 12/Dec/19

s i n (2 x) = c o s x \Leftrightarrow s i n (2 x) = s i n (\frac{π}{2} - x) \Leftrightarrow 2 x = \frac{π}{2} - x + 2 k π o r

2 x = \frac{π}{2} + x + 2 k π (k f r o m Z) \Leftrightarrow 3 x = \frac{π}{2} + 2 k π o r x = \frac{π}{2} + 2 k π \Leftrightarrow

x = \frac{π}{6} + \frac{2 k π}{3} o r x = \frac{π}{2} + 2 k π (k \in Z)

Commented by mathmax by abdo last updated on 12/Dec/19

c a s e 1 - π < \frac{π}{2} + 2 k π < π \Rightarrow - 1 < \frac{1}{2} + 2 k < 1 \Rightarrow - \frac{3}{2} < 2 k < \frac{1}{2} \Rightarrow

- \frac{3}{4} < k < \frac{1}{4} \Rightarrow - 0, \dots < k < 0, \dots \Rightarrow k = 0 \Rightarrow x = \frac{π}{2}

c a s e 2 - π < \frac{π}{6} + \frac{2 k π}{3} < π \Rightarrow - 1 < \frac{1}{6} + \frac{2 k}{3} < 1 \Rightarrow

- \frac{7}{6} < \frac{2 k}{3} < \frac{5}{6} \Rightarrow - \frac{7}{2} < 2 k < \frac{5}{2} \Rightarrow - \frac{7}{4} < k < \frac{5}{4} \Rightarrow - 1, \dots < k < 1, . . \Rightarrow

k \in {- 1, 0, 1}

k = - 1 \Rightarrow x = \frac{π}{6} - \frac{2 π}{3} = \frac{- 3 π}{6} = - \frac{π}{2}

k = 0 \Rightarrow x = \frac{π}{6}

k = 1 \Rightarrow x = \frac{π}{6} + \frac{2 π}{3} = \frac{5 π}{6}