a)căn 3 sin4x-cos4x-2cosx=0
b)cosx +căn 3 cos2x-căn 3 sinx-sin2x=0
c)cos 3x+sin2x=căn 3(sin3x+cos2x)
d)cosx +căn 3=3-3/cosx+căn 3 sinx+1
Những câu hỏi liên quan
Giúp mình với mn...
1)cos2x+cos22x+cos23x+cos24x=2
2) (1-tanx) (1+sin2x)=1+tanx
3) tan2x=sin3x.cosx
4) tanx +cot2x=2cot4x
5) sinx+sin2x+sin3x=cosx+cos2x+cos3x
6)sinx=√2 sin5x-cosx
7) 1/sin2x + 1/cos2x =2/sin4x
8) sinx+cosx=cos2x/1-sin2x
9)1+cos2x/cosx= sin2x/1-cos2x
10)sin3x+cos3x/2cosx-sinx=cos2x
1. sin^8(x) - cos^8(x) - 4sin^6(x) + 6sin^4(x) - 4sin^2(x) = 1
2. sin6x+sin4x+sin2x/1+cos2x+cos4x = 2sin2x
3. 1+sin2x /cosx+sinx - 1-tan^2(x/2)/1+tan^2(x/2) = sinx
4. cos4x + 4cos2x + 3 = 8cos^4(x)
5. 1+cosx+cos2x+cos3x/ 2cos^2(x)+cosx-1 = 2cosx
1. sin^8(x) - cos^8(x) - 4sin^6(x) + 6sin^4(x) - 4sin^2(x) = 1
2. sin6x+sin4x+sin2x/1+cos2x+cos4x = 2sin2x
3. 1+sin2x /cosx+sinx - 1-tan^2(x/2)/1+tan^2(x/2) = sinx
4. cos4x + 4cos2x + 3 = 8cos^4(x)
5. 1+cosx+cos2x+cos3x/ 2cos^2(x)+cosx-1 = 2cosx
\(sin^8x-cos^8x-4sin^6x+6sin^4x-4sin^2x\)
\(=sin^8x-\left(1-sin^2x\right)^4-4sin^6x+6sin^4x-4sin^2x\)
\(=sin^8x-\left(1-4sin^2x+6sin^4x-4sin^6x+sin^8x\right)-4sin^6x+6sin^4x-4sin^2x\)\(=-1\) (bạn chép nhầm đề)
b/ \(\frac{sin6x+sin2x+sin4x}{1+cos2x+cos4x}=\frac{2sin4x.cos2x+sin4x}{1+cos2x+2cos^22x-1}=\frac{sin4x\left(2cos2x+1\right)}{cos2x\left(2cos2x+1\right)}=\frac{sin4x}{cos2x}=\frac{2sin2x.cos2x}{cos2x}=2sin2x\)
c/ \(\frac{1+sin2x}{cosx+sinx}-\frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}=\frac{sin^2x+cos^2x+2sinx.cosx}{cosx+sinx}-\left(1-tan^2\frac{x}{2}\right)cos^2\frac{x}{2}\)
\(=\frac{\left(sinx+cosx\right)^2}{sinx+cosx}-\left(cos^2\frac{x}{2}-sin^2\frac{x}{2}\right)=sinx+cosx-cosx=sinx\)
d/ \(cos4x+4cos2x+3=2cos^22x-1+4cos2x+3\)
\(=2\left(cos^22x+2cos2x+1\right)=2\left(cos2x+1\right)^2=2\left(2cos^2x-1+1\right)^2=8cos^4x\)
e/
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tìm tập xác định
a)y=tan(pi/2 nhân cosx) b) y= cosx+1/cosx c) y=tan2xcot8x d)y=căn bậc hai của (2cosx-căn bậc hai của 3) e) y=(2+3sin2x)/cos2x-1 f)y=3sin3x/căn bậc hai (1-cosx) g)y=căn bậc hai của (2+3tan^22x) h) y=1/ căn bậc hai ( 1+sin^3x) k)y=sinx/ tan^2x/2
Bạn chú ý gõ đề bằng công thức toán (hộp biểu tượng $\sum$) trên thanh công cụ. Nhìn đề rối mắt thế này thật tình không ai muốn đọc chứ đừng nói đến giúp =)))
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cosx-2cos3x1+sqrt{3}sinx
sinx+sinxleft(x+dfrac{pi}{3}right)+sin4xsinleft(2x-dfrac{pi}{3}right)
left(1-dfrac{1}{2sinx}right)cos^22x2sinx-3+dfrac{1}{sinx}
( sinx -2cosx)cos2x + sinx (cos4x - 1)cosx +dfrac{cos2x}{2sinx}
left(dfrac{cos4x+sin2x}{cos3x+sin3x}right)^22sqrt{2}sinleft(x+dfrac{pi}{4}right)+3
Đọc tiếp
\(cosx-2cos3x=1+\sqrt{3}sinx\)
\(sinx+sinx\left(x+\dfrac{\pi}{3}\right)+sin4x=sin\left(2x-\dfrac{\pi}{3}\right)\)
\(\left(1-\dfrac{1}{2sinx}\right)cos^22x=2sinx-3+\dfrac{1}{sinx}\)
( sinx -2cosx)cos2x + sinx = (cos4x - 1)cosx +\(\dfrac{cos2x}{2sinx}\)
\(\left(\dfrac{cos4x+sin2x}{cos3x+sin3x}\right)^2=2\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)+3\)
Giải pt
\(2sin\left(x+\dfrac{\pi}{6}\right)+sinx+2cosx=3\)
\(\left(sin2x+cos2x\right)cosx+2cos2x-sinx=0\)
\(sin2x-cos2x+3sinx-cosx-1=0\)
1.
\(2sin\left(x+\dfrac{\pi}{6}\right)+sinx+2cosx=3\)
\(\Leftrightarrow\sqrt{3}sinx+cosx+sinx+2cosx=3\)
\(\Leftrightarrow\left(\sqrt{3}+1\right)sinx+3cosx=3\)
\(\Leftrightarrow\sqrt{13+2\sqrt{3}}\left[\dfrac{\sqrt{3}+1}{\sqrt{13+2\sqrt{3}}}sinx+\dfrac{3}{\sqrt{13+2\sqrt{3}}}cosx\right]=3\)
Đặt \(\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)
\(pt\Leftrightarrow\sqrt{13+2\sqrt{3}}sin\left(x+\alpha\right)=3\)
\(\Leftrightarrow sin\left(x+\alpha\right)=\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\\x+\alpha=\pi-arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)
Vậy phương trình đã cho có nghiệm:
\(x=k2\pi;x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\)
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2.
\(\left(sin2x+cos2x\right)cosx+2cos2x-sinx=0\)
\(\Leftrightarrow2sinx.cos^2x+cos2x.cosx+2cos2x-sinx=0\)
\(\Leftrightarrow\left(2cos^2x-1\right)sinx+cos2x.cosx+2cos2x=0\)
\(\Leftrightarrow cos2x.sinx+cos2x.cosx+2cos2x=0\)
\(\Leftrightarrow cos2x.\left(sinx+cosx+2\right)=0\)
\(\Leftrightarrow cos2x=0\)
\(\Leftrightarrow2x=\dfrac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)
Vậy phương trình đã cho có nghiệm \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)
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1) 2sinx + cosx = sin2x + 1
2) (1 + cosx)(1+sinx) = 2
3) 3cos4x - 8cos6x + 2cos2x +3 =0
4) sin3x + cos3x.sinx + cosx = \(\sqrt{2}\)cos2x
5) (2cosx -1)(2sinx + cosx) = sin2x - sinx
31 tháng 10 2023 lúc 20:08
\(sinx+4cosx=2+sin2x\)
\(\left(1-sin2x\right)\left(sinx+cosx\right)=cos2x\)
\(1+sinx+cosx+sin2x+cos2x=0\)
\(sinx+sin2x+sin3x=1+cosx+cos2x\)
\(sin^22x-cos^28x=sin\left(\dfrac{17\pi}{2}+10x\right)\)
1> 1 + sinx + cosx + sin2x + cos2x = 0
2> cos2x + 3sin2x + 5 sinx - 3cosx = 3
3> \(\dfrac{\sqrt{2}*(cosx - sinx)}{cotx - 1}\) = \(\dfrac{1}{tanx + cot2x}\)
4> (2cosx - 1)*(2sinx + cosx) = sin2x - sinx