Question

розвязати рівняння sin^2 (x) + sin^2 (3x) = 1

Answer 1

$\displaystyle\bf\\Sin^{2} x+Sin^{2} 3x=1\\\\\\\frac{1-Cos2x}{2} +\frac{1-Cos6x}{2}=1\\\\\\\frac{1}{2} -\frac{1}{2}Cos2x+ \frac{1}{2} -\frac{1}{2}Cos6x=1\\\\\\1 -\frac{1}{2}Cos2x -\frac{1}{2}Cos6x=1\\\\\\ -\frac{1}{2}Cos2x -\frac{1}{2}Cos6x=0\\\\\\Cos2x+Cos6x=0\\\\\\2Cos\frac{2x+6x}{2} Cos\frac{2x-6x}{2} =0\\\\\\Cos4x\cdot Cos2x=0\\\\\\\left[\begin{array}{ccc}Cos4x=0\\Cos2x=0\end{array}\right$

$\displaystyle\bf\\\left[\begin{array}{ccc}4x=\dfrac{\pi }{2} +\pi n,n\in Z\\2x=\dfrac{\pi }{2}+\pi n,n\in Z \end{array}\right \\\\\\\left \{ {{x=\dfrac{\pi }{8} +\dfrac{\pi n}{4},n\in Z } \atop {x=\dfrac{\pi }{4}+\dfrac{\pi n}{2},n\in Z }} \right.$