Phương trình lượng giác

Question

1)

$\frac{4sinx.sin\left ( x + \frac{\Pi }{3}\right )+5\sqrt{3}sinx+3(cos+2)}{1-2cosx}=1$

2)

$5sin3\left ( \frac{\Pi }{3}+x \right )+3sin5\left ( \frac{\Pi }{5}-x \right )=0$

3)

$\frac{cos^{3}x-cos^{2}x}{sinx+cosx}=2(1+sinx)$

4)

$2(sinx-cosx)+sin3x+cos3x=3\sqrt{2}(2+sin2x)$

5)

$2sin2x+\sqrt{2}sin\left ( 2x+\frac{\Pi }{4} \right )+11sinx-3cosx=6$

6)

$2sinx(cos^{2}x-sin^{2}x)=sinx+\sqrt{3}cos3x$

7)

$1+cot2x=\frac{4sin^{2}x}{1-cos4x}$

8)

$\frac{\sqrt{3}}{2cos^{2}x}+\frac{2+sin2x}{sin2x}=\frac{1}{tanx}+1+\sqrt{3}$

tran85295 · Answer 1 · 8/24/2016 6:52:50 PM

7)ĐKXĐ

$x\ne \frac{k\pi}{2}$

$pt\Leftrightarrow 1+\frac{1-\tan^2x}{\tan x}=\frac{4\sin^2x}{\sin^22x}$

$\Leftrightarrow 1+\frac{1-\tan^2x}{\tan x}=\frac{1}{2\cos^2x}$

$\Leftrightarrow 2\left(1+\frac{1-\tan^2x}{\tan x} \right)=1+\tan^2x$

$\Leftrightarrow \left[ \begin{array}{l} \tan x=1\\ \tan x=-1 \end{array} \right.\Leftrightarrow x=\frac{k\pi}{2}-\frac{\pi}{4}$

6)

$pt\Leftrightarrow 2\sin x\cos 2x=\sin x+\sqrt 3\cos 3x$

$\Leftrightarrow \sin 3x-\sqrt 3 \cos 3x=2\sin x$

$\Leftrightarrow -2\cos \left(3x+\frac{\pi}6\right)=2\sin x$

$\Leftrightarrow \cos \left(3x+\frac{\pi}6 \right)=\cos \left( x+\frac{\pi}{2}\right)$

1 Đáp án