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Question

$2(1+cosx) = \sqrt{3}tg(\frac{\pi}{2} -\frac{x}{2})$

score 1 · Answer 1

$2(1+\cos x)=\sqrt{3}\tan (\frac{\pi}{2}-\frac{x}{2})\\\Leftrightarrow 2(1+\cos x)=\sqrt{3}\cot \frac{x}{2}\\\Leftrightarrow 2(1+2\cos^{2}\frac{x}{2}-1)-\sqrt{3}\frac{\cos \frac{x}{2}}{\sin\frac{x}{2}}=0\\\Leftrightarrow 4\cos^{2}\frac{x}{2}\sin\frac{x}{2}-\sqrt{3}\cos \frac{x}{2}=0\\\Leftrightarrow \cos\frac{x}{2}(4\cos\frac{x}{2}\sin\frac{x}{2}-\sqrt{3})=0\\\Leftrightarrow \cos\frac{x}{2}(2\sin x-\sqrt{3})=0\\\Leftrightarrow \cos \frac{x}{2}=0 (1)\\hoặc \sin x=\frac{\sqrt{3}}{2} (2)$
Sau đó giải pt (1), (2) ra nghiệm.

score 1 · Answer 2

Bình chọn tăng 1 Bình chọn giảm

1. 2cos2(x−270)+5sin(x−90)=4
Ta có : $\cos ^{2}(x-270^{o})=(\cos x\cos270^{o}+\sin x\sin270^{o})^{2}=(-\sin x)^{2}=\sin^{2}x$
$\sin (x-90^{o})=\sin x\cos90^{o}-\cos x\sin90^{o}=-\cos x$
PT$\Leftrightarrow 2\sin^{2}x-5\cos x-4=0$
$\Leftrightarrow 2(1-\cos^{2}x) -5\cos x-4=0$
$\Leftrightarrow 2\cos ^{2}x+5\cos x+2=0$
$\Leftrightarrow \cos x=\frac{-1}{2} (thỏa mãn) \\hoặc\cos x=-2 (loại) \\ \Leftrightarrow x=\pm 120^{o}+k360^{o}$