Bài 101911

Question

Giải các phương trình :
a) $6\sin ^2x+7\sqrt{3}\sin 2x-8\cos^2 x-6=0$
b) $ 4\cos^ 2\frac{x}{2}+\frac{1}{2}\sin x+3\sin^2 \frac{x}{2}=3$

score 0 · Answer 1

a) PT $
\Leftrightarrow       7\sqrt{3}\sin 2x-14\cos^2 x=0
$

             $ \Leftrightarrow        14\cos x(\sqrt{3}\sin x-\cos x)=0$

             $\Leftrightarrow      \left[ \begin{array}{l}\cos x = 0\\\tan x = \frac{1}{\sqrt{3}}\end{array} \right. \Leftrightarrow      \left[ \begin{array}{l}x = \frac{\pi}{2}+k\pi\\x = \frac{\pi}{6}+k\pi\end{array} \right.              (k\in Z)$

b) PT    $\Leftrightarrow      2(1+\cos x)+\frac{1}{2}\sin x+\frac{3}{2}(1-\cos x)=3$

              $\Leftrightarrow      4+4\cos x+\sin x+3-3\cos x=6$

              $\Leftrightarrow     \sin x+\cos x=-1 \Leftrightarrow \sqrt{2}\sin (x+\frac{\pi}{4})=-1$

              $\Leftrightarrow \sin(x+\frac{\pi}{4})=-\frac{\sqrt{2}}{2}=\sin(-\frac{\pi}{4})$

               $\Leftrightarrow     \left[ \begin{array}{l}x+\frac{\pi}{4} = -\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4} = \pi+\frac{\pi}{4}+k2\pi\end{array} \right. \Leftrightarrow       \left[ \begin{array}{l}x = -\frac{\pi}{2}+k2\pi\\x =\pi+k2\pi \end{array} \right.                  (k\in Z)$

Bài 101911

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Bài 101911

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