Question

Решите уравнение sin2x = 2sin^2(π/2− x)

Answer 1

$Sin2x=2Sin^{2}\Big(\dfrac{\pi }{2}-x\Big)\\\\Sin2x=2Cos^{2}x\\\\2Sinx Cosx-2Cos^{2}x=0\\\\2Cosx(Sinx-Cosx)=0\\\\\left[\begin{array}{ccc}Cosx=0\\Sinx-Cosx=0\end{array}\right\\\\\\\left[\begin{array}{ccc}x=\dfrac{\pi }{2}+\pi n,n\in Z \\\dfrac{Sinx}{Sinx }-\dfrac{Cosx}{Sinx}=0 \ , \ Sinx\neq 0\end{array}\right$

$\left[\begin{array}{ccc}x=\dfrac{\pi }{2}+\pi n,n\in Z \\1-Ctgx=0\end{array}\right\\\\\\\left[\begin{array}{ccc}x=\dfrac{\pi }{2}+\pi n,n\in Z \\Ctgx=1\end{array}\right\\\\\\\left[\begin{array}{ccc}x=\dfrac{\pi }{2}+\pi n,n\in Z \\x=\dfrac{\pi }{4}+\pi n,n\in Z \end{array}\right$

Answer 2

пункт 1

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