Toán sin ²2x-2 /sin ²2x -4cos ² x =tan ² x 08/09/2021 By Kennedy sin ²2x-2 /sin ²2x -4cos ² x =tan ² x
Đáp án: x=π4+kπ2(k∈Z)x=π4+kπ2(k∈Z) Giải thích các bước giải: sin22x−24cos2x.(sin2x−1)=sin2xcos2xsin22x−2−4cos2x=sin2xsin22x−2=−4sin2x.cos2x=−sin22x2sin22x−2=0sin22x=1[sin2x=1sin2x=−1↔[2x=π2+k2π2x=−π2+k2π↔2x=π2+kπ↔x=π4+kπ2(k∈Z)(tm) Trả lời
Đáp án: \(x = \frac{\pi }{4} + \frac{{k\pi }}{2}(k \in Z)\) Giải thích các bước giải: \(\begin{array}{l}\frac{{{{\sin }^2}2x – 2}}{{4{{\cos }^2}x.({\mathop{\rm s}\nolimits} {\rm{i}}{{\rm{n}}^2}x – 1)}} = \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}\\\frac{{{{\sin }^2}2x – 2}}{{ – 4{{\cos }^2}x}} = {\sin ^2}x\\{\sin ^2}2x – 2 = – 4{\sin ^2}x.{\cos ^2}x = – {\sin ^2}2x\\2{\sin ^2}2x – 2 = 0\\{\sin ^2}2x = 1\\\left[ \begin{array}{l}\sin 2x = 1\\\sin 2x = – 1\end{array} \right. \leftrightarrow \left[ \begin{array}{l}2x = \frac{\pi }{2} + k2\pi \\2x = \frac{{ – \pi }}{2} + k2\pi \end{array} \right. \leftrightarrow 2x = \frac{\pi }{2} + k\pi \leftrightarrow x = \frac{\pi }{4} + \frac{{k\pi }}{2}(k \in Z)(tm)\end{array}\) Trả lời
Đáp án:
x=π4+kπ2(k∈Z)x=π4+kπ2(k∈Z)
Giải thích các bước giải:
sin22x−24cos2x.(sin2x−1)=sin2xcos2xsin22x−2−4cos2x=sin2xsin22x−2=−4sin2x.cos2x=−sin22x2sin22x−2=0sin22x=1[sin2x=1sin2x=−1↔[2x=π2+k2π2x=−π2+k2π↔2x=π2+kπ↔x=π4+kπ2(k∈Z)(tm)
Đáp án:
\(x = \frac{\pi }{4} + \frac{{k\pi }}{2}(k \in Z)\)
Giải thích các bước giải:
\(\begin{array}{l}
\frac{{{{\sin }^2}2x – 2}}{{4{{\cos }^2}x.({\mathop{\rm s}\nolimits} {\rm{i}}{{\rm{n}}^2}x – 1)}} = \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}\\
\frac{{{{\sin }^2}2x – 2}}{{ – 4{{\cos }^2}x}} = {\sin ^2}x\\
{\sin ^2}2x – 2 = – 4{\sin ^2}x.{\cos ^2}x = – {\sin ^2}2x\\
2{\sin ^2}2x – 2 = 0\\
{\sin ^2}2x = 1\\
\left[ \begin{array}{l}
\sin 2x = 1\\
\sin 2x = – 1
\end{array} \right. \leftrightarrow \left[ \begin{array}{l}
2x = \frac{\pi }{2} + k2\pi \\
2x = \frac{{ – \pi }}{2} + k2\pi
\end{array} \right. \leftrightarrow 2x = \frac{\pi }{2} + k\pi \leftrightarrow x = \frac{\pi }{4} + \frac{{k\pi }}{2}(k \in Z)(tm)
\end{array}\)