Giải phương trình: $\sin x+\sqrt{3}\cos x=\sqrt{2+\cos2x+\sqrt{3}\sin2x } $
|
a) Tìm tất cả các nghiệm nguyên của phương trình sau:$\cos \left[ {\frac{\pi }{8}\left( {3x - \sqrt {9{x^2} + 160x + 800} } \right)} \right] = 1$b) Biết rằng $\sin 15^0= \frac{\sqrt{6 } -\sqrt{2 } }{4}$. Tính các tỉ số lượng giác của góc...
|
Giải phương trình $$ \sin^{7}x + \cos^{16}x =1$$
Trả lời 08-09-17 09:01 AM
|
giải phương trình: sin^2(x) + sinx + 2cosx =2
|
Giải phương trình : $\sin 2x+\sin x-\frac{1}{2\sin x}-\frac{1}{\sin 2x} =2\cot 2x $
|
1. $ tan^{2}x + cot^{2}x+2(tanx+cotx)=6$ 2. $ 4(sin^{2}x+\frac{1}{sin^{2}x})+4(sinx+\frac{1}{sinx})-7=0$
Trả lời 26-07-16 06:03 PM
|
1. $ tan^{2}x + cot^{2}x+2(tanx+cotx)=6$ 2. $ 4(sin^{2}x+\frac{1}{sin^{2}x})+4(sinx+\frac{1}{sinx})-7=0$
Trả lời 26-07-16 03:59 PM
|
1. $ tan^{2}x + cot^{2}x+2(tanx+cotx)=6$ 2. $ 4(sin^{2}x+\frac{1}{sin^{2}x})+4(sinx+\frac{1}{sinx})-7=0$
Trả lời 26-07-16 03:58 PM
|
1) $sin^6x$ - $cos^6x$ + 2cos2x - $cos^24x$ =1 2) $sin^6x$ - $cos^6x$ = 13/12
Trả lời 19-07-16 04:24 PM
|
1) $sin^6x$ - $cos^6x$ + 2cos2x - $cos^24x$ =1 2) $sin^6x$ - $cos^6x$ = 13/12
Trả lời 19-07-16 03:55 PM
|
1.$ (1+sin^{2}x)cosx+(1+cos^{2})sinx=1+sin2x$2.$tan^{2}x+cot^{2}x +2(tanx+cotx)=6$3.$4(sin^{2}x+\frac{1}{sin^{2}x}) +4(sinx+\frac{1}{sinx}) - 7=0 $4.$\frac{3}{cos^{2}x}+3cot^{2}x+4.(tanx+cotx)-1=0 $
Trả lời 19-07-16 03:46 PM
|
$3\tan^2x+4\sin^2 x-2\sqrt{3}\tan x-4\sin x+2=0$$\cos^5 x+x^2=0$$\cos x=1-\frac{x^2}{2}\forall x>0$$\sin^4 x+\cos^{15} x=1$ , $2\sin^3 x+ 4\cos^3 x=3\sin x$ ,$1+\cos^2 x=\sin 2x+\sin 4x$2sin3x+4cos3x=3sinx" role="presentation" style="display:...
Trả lời 18-07-16 04:02 PM
|
$3\tan^2x+4\sin^2 x-2\sqrt{3}\tan x-4\sin x+2=0$$\cos^5 x+x^2=0$$\cos x=1-\frac{x^2}{2}\forall x>0$$\sin^4 x+\cos^{15} x=1$ , $2\sin^3 x+ 4\cos^3 x=3\sin x$ ,$1+\cos^2 x=\sin 2x+\sin 4x$2sin3x+4cos3x=3sinx" role="presentation" style="display:...
Trả lời 18-07-16 04:01 PM
|
$3\tan^2x+4\sin^2 x-2\sqrt{3}\tan x-4\sin x+2=0$$\cos^5 x+x^2=0$$\cos x=1-\frac{x^2}{2}\forall x>0$$\sin^4 x+\cos^{15} x=1$ , $2\sin^3 x+ 4\cos^3 x=3\sin x$ ,$1+\cos^2 x=\sin 2x+\sin 4x$2sin3x+4cos3x=3sinx" role="presentation" style="display:...
Trả lời 18-07-16 03:23 PM
|
$3\tan^2x+4\sin^2 x-2\sqrt{3}\tan x-4\sin x+2=0$$\cos^5 x+x^2=0$$\cos x=1-\frac{x^2}{2}\forall x>0$$\sin^4 x+\cos^{15} x=1$ , $2\sin^3 x+ 4\cos^3 x=3\sin x$ ,$1+\cos^2 x=\sin 2x+\sin 4x$2sin3x+4cos3x=3sinx" role="presentation" style="display:...
Trả lời 18-07-16 03:10 PM
|
$3\tan^2x+4\sin^2 x-2\sqrt{3}\tan x-4\sin x+2=0$$\cos^5 x+x^2=0$$\cos x=1-\frac{x^2}{2}\forall x>0$$\sin^4 x+\cos^{15} x=1$ , $2\sin^3 x+ 4\cos^3 x=3\sin x$ ,$1+\cos^2 x=\sin 2x+\sin 4x$2sin3x+4cos3x=3sinx" role="presentation" style="display:...
Trả lời 18-07-16 02:57 PM
|
$3\tan^2x+4\sin^2 x-2\sqrt{3}\tan x-4\sin x+2=0$$\cos^5 x+x^2=0$$\cos x=1-\frac{x^2}{2}\forall x>0$$\sin^4 x+\cos^{15} x=1$ , $2\sin^3 x+ 4\cos^3 x=3\sin x$ ,$1+\cos^2 x=\sin 2x+\sin 4x$2sin3x+4cos3x=3sinx" role="presentation" style="display:...
Trả lời 18-07-16 02:49 PM
|
$\cos^2 x+\sqrt{3}\sin 2x=\sin^3 x+1$$\sin x+\cos x=\tan x+1$$1+\sin 2x=\sin 3x$$8\cos^3 x-2\cos 2x-4\cos x-1=0$$3\cos 4x-8\cos^6 x+2\cos ^2x+3=0$
Trả lời 18-07-16 02:41 PM
|
$\cos^2 x+\sqrt{3}\sin 2x=\sin^3 x+1$$\sin x+\cos x=\tan x+1$$1+\sin 2x=\sin 3x$$8\cos^3 x-2\cos 2x-4\cos x-1=0$$3\cos 4x-8\cos^6 x+2\cos ^2x+3=0$
Trả lời 18-07-16 11:02 AM
|
$\cos^2 x+\sqrt{3}\sin 2x=\sin^3 x+1$$\sin x+\cos x=\tan x+1$$1+\sin 2x=\sin 3x$$8\cos^3 x-2\cos 2x-4\cos x-1=0$$3\cos 4x-8\cos^6 x+2\cos ^2x+3=0$
Trả lời 18-07-16 10:18 AM
|