giúp vs ak

Question

$1) \frac{sinx+sin5x}{cos2x}=\sqrt{2}(sin2x+cos2x)$

$2) \sqrt{3}cos2x+sin2x=\frac{1}{2}(tanx+cotx)$

$3) 4cosx.cos3x+\sqrt{3}sin2x+2cos^{2}x$

tran85295 · Answer 1 · 7/12/2016 1:56:47 AM

1) Đk

$\cos 2x \ne0$

$pt\Leftrightarrow \frac{2\sin 3x.\cos 2x}{\cos 2x}=\sqrt 2.\sqrt 2\sin \left(2x+\frac{\pi}4 \right)$

$\Leftrightarrow \sin 3x=\sin \left(2x+\frac{\pi}4 \right)$

$\Leftrightarrow \left[ \begin{array}{l} x= \dfrac {\pi}4+k2\pi \\ x=\dfrac{3\pi}{20}+ \dfrac{k2\pi}{5} \end{array} \right. \quad(k\in \mathbb{Z})$

2) Đk

$\sin x \ne0, \cos x \ne0$

$pt\Leftrightarrow \sqrt 3 \cos 2x+\sin 2x=\frac{\sin^2x+\cos^2x}{2\sin x \cos x}$

$\Leftrightarrow \sqrt 3 \cos 2x+\sin 2x= \frac 1{\sin 2x}$

$\Leftrightarrow \sqrt 3.\sin 4x+2\sin^22x=2$

$\Leftrightarrow \sqrt 3 \sin 4x+1-\cos 4x=2$

$\Leftrightarrow \sqrt 3 \sin 4x-\cos 4x=1$

$\Leftrightarrow \sin \left( \frac{\pi}6 -4x \right)=-\frac 12$

$\Leftrightarrow \left[ \begin{array}{l} x=\dfrac{12k\pi-\pi}{6}\\x= \dfrac{-2k\pi-\pi}{4} \end{array} \right.\quad ( k \in \mathbb{Z})$

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