Question

giai pt

Answer 1

ta có sin²(2x)=4sin²xcos²x=4sin²x(1-sin²x)=4sin²x-4sin⁴x

sin²(3x)=16sin⁶x-24sin⁴x+9sin²x=>16sin⁶x-24sin⁴x+9sin²x+4sin²x-4sin⁴x=6

Đặt sin²x=a ta có 16a³-28a²+13a=6

giải tiếp....

Answer 2

sin²x + sin²2x + sin²3x = 2

⇔ sin²2x = 1 - sin²x + 1 - sin²3x

⇔ sin²2x = cos²x + cos²3x

⇔ \(\dfrac{1-cos4x}{2}=\dfrac{1+cos2x}{2}+\dfrac{1+cos6x}{2}\)

⇔ 1 - cos4x = 2 + cos2x + cos6x

⇔ cos2x + cos6x + cos4x + 1 = 0

⇔ 2cos4x . cos2x + 2cos²2x = 0

⇔ \(\left[{}\begin{matrix}cos2x=0\\cos4x+cos2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\2cos3x.cosx=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}cos2x=0\\cos3x=0\end{matrix}\right.\). Còn lại tự giải nhé !!