Question

Как решить уравнение:
(sinx)^2–(sin2x)^2+(sin3x)^2=1/2

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Answer 1

$(Sinx)^{2} -(Sin2x)^{2}+(Sin3x)^{2}=\frac{1}{2}\\\\\frac{1-Cos2x}{2}-\frac{1-Cos4x}{2}+\frac{1-Cos6x}{2}=\frac{1}{2}|*2\\\\(1-Cos2x)-(1-Cos4x)+(1-Cos6x)=1\\\\1-Cos2x-1+Cos4x+1-Cos6x=1\\\\-Cos2x+Cos4x-Cos6x=0\\\\Cos4x-(Cos2x+Cos6x)=0\\\\Cos4x-2Cos\frac{2x+6x}{2}Cos\frac{2x-6x}{2}=0\\\\Cos4x-2Cos4xCos2x=0\\\\Cos4x(1-2Cos2x)=0$

$\left[\begin{array}{ccc}Cos4x=0\\Cos2x=\frac{1}{2} \end{array}\right\\\\\\\left[\begin{array}{ccc}4x=\frac{\pi }{2}+\pi n,n\in Z  \\2x=\pm\frac{\pi }{3}+2\pi n,n\in Z  \end{array}\right\\\\\\\left[\begin{array}{ccc}x=\frac{\pi }{8}+\frac{\pi n }{4},n\in Z  \\x=\pm\frac{\pi }{6}+\pi n,n\in Z  \end{array}\right\\\\Otvet:\boxed{\frac{\pi }{8}+\frac{\pi n }{4};\pm\frac{\pi }{6}+\pi n,n\in Z}$