Toán giúp mình với: Cos2x (1+2Cos2x) = 1+Sinx(1-2Sin3x)+ $\sqrt[]{3}$Cosx. 06/10/2021 By Mackenzie giúp mình với: Cos2x (1+2Cos2x) = 1+Sinx(1-2Sin3x)+ $\sqrt[]{3}$Cosx.
Đáp án: \[\left[ \begin{array}{l}x = – \dfrac{\pi }{6} + k2\pi \\x = \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}\end{array} \right.\,\,\,\,\,\,\,\,\,\left( {k \in Z} \right)\] Giải thích các bước giải: Ta có: \(\begin{array}{l} – 2\sin x.\sin y = \cos \left( {x + y} \right) – \cos \left( {x – y} \right)\\\cos 2x\left( {1 + 2\cos 2x} \right) = 1 + \sin x\left( {1 – 2\sin 3x} \right) + \sqrt 3 \cos x\\ \Leftrightarrow \cos 2x + 2{\cos ^2}2x = 1 + \sin x – 2\sin x.\sin 3x + \sqrt 3 \cos x\\ \Leftrightarrow \cos 2x + \left( {2{{\cos }^2}2x – 1} \right) = \sin x + \cos \left( {x + 3x} \right) – \cos \left( {x – 3x} \right) + \sqrt 3 \cos x\\ \Leftrightarrow \cos 2x + \cos 4x = \sin x + \cos 4x – \cos \left( { – 2x} \right) + \sqrt 3 \cos x\\ \Leftrightarrow \sin x + \sqrt 3 \cos x = 2\cos 2x\\ \Leftrightarrow \dfrac{1}{2}\sin x + \dfrac{{\sqrt 3 }}{2}\cos x = \cos 2x\\ \Leftrightarrow \cos x.\cos \dfrac{\pi }{6} + \sin x.\sin \dfrac{\pi }{6} = \cos 2x\\ \Leftrightarrow \cos \left( {x – \dfrac{\pi }{6}} \right) = \cos 2x\\ \Leftrightarrow \left[ \begin{array}{l}x – \dfrac{\pi }{6} = 2x + k2\pi \\x – \dfrac{\pi }{6} = – 2x + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = – \dfrac{\pi }{6} + k2\pi \\x = \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}\end{array} \right.\,\,\,\,\,\,\,\,\,\left( {k \in Z} \right)\end{array}\) Vậy nghiệm của phương trình đã cho là \(\left[ \begin{array}{l}x = – \dfrac{\pi }{6} + k2\pi \\x = \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}\end{array} \right.\,\,\,\,\,\,\,\,\,\left( {k \in Z} \right)\) Trả lời
Đáp án: $x = – \frac{π}{6} + k2π; x = \frac{π}{18} + k\frac{2π}{3}$ Giải thích các bước giải: $cos2x(1 + 2cos2x) = 1 + sinx(1- 2sin3x) + \sqrt[]{3}.cosx$ $⇔ cos2x + (2cos²2x – 1) = sinx – 2sin3xsinx + \sqrt[]{3}.cosx$ $⇔ cos2x + cos4x = sinx + (cos4x – cos2x) + \sqrt[]{3}.cosx$ $⇔ 2cos2x = \sqrt[]{3}.cosx + sinx$ $⇔ cos2x = \frac{\sqrt[]{3}}{2}.cosx + \frac{1}{2}.sinx$ $⇔ cos2x = cos\frac{π}{6}.cosx + sin\frac{π}{6}.sinx$ $⇔ cos2x = cos(x – \frac{π}{6})$ @ $2x = x – \frac{π}{6} + k2π ⇔ x = – \frac{π}{6} + k2π $ @ $2x = – (x – \frac{π}{6}) + k2π ⇔ x = \frac{π}{18} + k\frac{2π}{3} $ Trả lời
Đáp án:
\[\left[ \begin{array}{l}
x = – \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}
\end{array} \right.\,\,\,\,\,\,\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
– 2\sin x.\sin y = \cos \left( {x + y} \right) – \cos \left( {x – y} \right)\\
\cos 2x\left( {1 + 2\cos 2x} \right) = 1 + \sin x\left( {1 – 2\sin 3x} \right) + \sqrt 3 \cos x\\
\Leftrightarrow \cos 2x + 2{\cos ^2}2x = 1 + \sin x – 2\sin x.\sin 3x + \sqrt 3 \cos x\\
\Leftrightarrow \cos 2x + \left( {2{{\cos }^2}2x – 1} \right) = \sin x + \cos \left( {x + 3x} \right) – \cos \left( {x – 3x} \right) + \sqrt 3 \cos x\\
\Leftrightarrow \cos 2x + \cos 4x = \sin x + \cos 4x – \cos \left( { – 2x} \right) + \sqrt 3 \cos x\\
\Leftrightarrow \sin x + \sqrt 3 \cos x = 2\cos 2x\\
\Leftrightarrow \dfrac{1}{2}\sin x + \dfrac{{\sqrt 3 }}{2}\cos x = \cos 2x\\
\Leftrightarrow \cos x.\cos \dfrac{\pi }{6} + \sin x.\sin \dfrac{\pi }{6} = \cos 2x\\
\Leftrightarrow \cos \left( {x – \dfrac{\pi }{6}} \right) = \cos 2x\\
\Leftrightarrow \left[ \begin{array}{l}
x – \dfrac{\pi }{6} = 2x + k2\pi \\
x – \dfrac{\pi }{6} = – 2x + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = – \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}
\end{array} \right.\,\,\,\,\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
Vậy nghiệm của phương trình đã cho là \(\left[ \begin{array}{l}
x = – \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}
\end{array} \right.\,\,\,\,\,\,\,\,\,\left( {k \in Z} \right)\)
Đáp án: $x = – \frac{π}{6} + k2π; x = \frac{π}{18} + k\frac{2π}{3}$
Giải thích các bước giải:
$cos2x(1 + 2cos2x) = 1 + sinx(1- 2sin3x) + \sqrt[]{3}.cosx$
$⇔ cos2x + (2cos²2x – 1) = sinx – 2sin3xsinx + \sqrt[]{3}.cosx$
$⇔ cos2x + cos4x = sinx + (cos4x – cos2x) + \sqrt[]{3}.cosx$
$⇔ 2cos2x = \sqrt[]{3}.cosx + sinx$
$⇔ cos2x = \frac{\sqrt[]{3}}{2}.cosx + \frac{1}{2}.sinx$
$⇔ cos2x = cos\frac{π}{6}.cosx + sin\frac{π}{6}.sinx$
$⇔ cos2x = cos(x – \frac{π}{6})$
@ $2x = x – \frac{π}{6} + k2π ⇔ x = – \frac{π}{6} + k2π $
@ $2x = – (x – \frac{π}{6}) + k2π ⇔ x = \frac{π}{18} + k\frac{2π}{3} $