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2. 3 tan ( x/2 -pi/3)= căn 3 3. căn 3 Sin x + Cos x = 2 Giúp mik vs cần gấp cảm ơn mọi ng nhìu
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Đáp án:
$2) \, x = \pi + k2\pi \quad (k \in \Bbb Z)$
$3) \, x = \dfrac{\pi}{3} + k2\pi\quad (k \in \Bbb Z)$
Giải thích các bước giải:
$2) \,\,3\tan\left(\dfrac{x}{2} – \dfrac{\pi}{3}\right) = \sqrt3\qquad (*)$
$ĐKXĐ: \, \cos\left(\dfrac{x}{2} – \dfrac{\pi}{3}\right) \ne 0 \Leftrightarrow \dfrac{5\pi}{3} + n2\pi \quad (n \in \Bbb Z)$
$(*)\Leftrightarrow \tan\left(\dfrac{x}{2} – \dfrac{\pi}{3}\right) = \dfrac{\sqrt3}{3}$
$\Leftrightarrow \tan\left(\dfrac{x}{2} – \dfrac{\pi}{3}\right) = \tan\dfrac{\pi}{6}$
$\Leftrightarrow \dfrac{x}{2} – \dfrac{\pi}{3} = \dfrac{\pi}{6} + k\pi$
$\Leftrightarrow \dfrac{x}{2} = \dfrac{\pi}{2} + k\pi$
$\Leftrightarrow x = \pi + k2\pi \quad (k \in \Bbb Z)$
$3)\,\, \sqrt3\sin x + \cos x = 2$
$\Leftrightarrow \dfrac{\sqrt3}{2}\sin x + \dfrac{1}{2}\cos x = 1$
$\Leftrightarrow \sin x\cos\dfrac{\pi}{6} + \cos x\sin\dfrac{\pi}{6} = 1$
$\Leftrightarrow \sin\left(x + \dfrac{\pi}{6}\right) = 1$
$\Leftrightarrow x + \dfrac{\pi}{6} = \dfrac{\pi}{2} + k\pi$
$\Leftrightarrow x = \dfrac{\pi}{3} + k2\pi\quad (k \in \Bbb Z)$
Giải thích các bước giải:
$\text{2) 3tan($\dfrac{x}{2}$ – $\dfrac{\pi}{3}$ )=$\sqrt[]{3}$ }$
⇔$\text{tan($\dfrac{x}{2}$ – $\dfrac{\pi}{3}$ )=$\dfrac{\sqrt[]{3}}{3}$ }$
⇔$\text{$\dfrac{x}{2}$ – $\dfrac{\pi}{3}$ = $\dfrac{\pi}{6}$ +k$\pi$}$
⇔$\text{$\dfrac{x}{2}$ = $\dfrac{\pi}{2}$ +k$\pi$}$
⇒$\text{${x}$ = $\pi$ +k2$\pi$ (k∈Z)}$
$\text{3) $\sqrt[]{3}$sinx + cosx = 2 }$
chia hai vế cho 2 ta được:
⇒$\text{ $\dfrac{\sqrt[]{3}}{2}$ sinx + $\dfrac{1}{2}$ cosx = 1 }$
⇔$\text{ sin(x + $\dfrac{\pi}{6}$)=1 }$
⇔$\text{ x + $\dfrac{\pi}{6}$= $\dfrac{\pi}{2}$ +k2$\pi$ }$
⇒$\text{ x = $\dfrac{\pi}{3}$ +k2$\pi$ (k∈Z) }$