Question

sin x/2 +cosx=1 help plizzz​

Answer 1

$\sin \dfrac{x}{2} + \cos x = 1\\\\\sin \dfrac{x}{2} + \cos x - 1 = 0\\\\\sin \dfrac{x}{2} +2\sin^{2}\dfrac{x}{2} = 0\\\\\sin \dfrac{x}{2} \bigg(1 + 2\sin \dfrac{x}{2} \bigg) = 0$

$\left[\begin{array}{ccc}\sin \dfrac{x}{2} = 0 \ \ \ \ \ \ \ \\ 1 + 2\sin \dfrac{x}{2} = 0 \end{array}\right$

$\left[\begin{array}{ccc}\sin \dfrac{x}{2} = 0 \ \ \  \\ \\ \sin \dfrac{x}{2} = -\dfrac{1}{2}  \end{array}\right$

$\left[\begin{array}{ccc}\dfrac{x}{2} = \pi n, \ n \in Z \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \\\dfrac{x}{2} = (-1)^{k} \bigg(-\dfrac{\pi}{6} \bigg) + \pi k, \ k \in Z \end{array}\right\\\\\\\left[\begin{array}{ccc}x = 2\pi n, \ n \in Z \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x = (-1)^{k+1} \dfrac{\pi}{3} + 2 \pi k, \ k \in Z \end{array}\right$

Ответ: $x = 2\pi n, \ n \in Z$, $x = (-1)^{k+1} \dfrac{\pi}{3} + 2 \pi k, \ k \in Z$