9 sin ^2 2x – 6 sin 2x+1=0 cot^2 3x+ 3 cot 3x+2+=0 cos4x+7cos2x- 3=0 cos2x+2cosx= 2sin^2 x/2 21/07/2021 Bởi Faith 9 sin ^2 2x – 6 sin 2x+1=0 cot^2 3x+ 3 cot 3x+2+=0 cos4x+7cos2x- 3=0 cos2x+2cosx= 2sin^2 x/2
Đáp án: 9sin²2x – 6sin2x + 1 =0 ⇔ (3 sin2x – 1)² =0 ⇔3sin2x – 1 =0 ⇔sin2x = 1/3 ⇔ \(\left[ \begin{array}{l}2x=arcsin\frac{1}{3}+ k2\pi \\2x=\pi -arcsin\frac{1}{3}+ k2\pi\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=arcsin\frac{1}{6}+ k\pi \\x=\pi/2 -arcsin\frac{1}{6}+ k\pi\end{array} \right.\) Bình luận
Đáp án:
9sin²2x – 6sin2x + 1 =0
⇔ (3 sin2x – 1)² =0
⇔3sin2x – 1 =0
⇔sin2x = 1/3
⇔ \(\left[ \begin{array}{l}2x=arcsin\frac{1}{3}+ k2\pi \\2x=\pi -arcsin\frac{1}{3}+ k2\pi\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=arcsin\frac{1}{6}+ k\pi \\x=\pi/2 -arcsin\frac{1}{6}+ k\pi\end{array} \right.\)