Do \(sin\left(a+k2\pi\right)=sina\)
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Bài 3: Một số phương trình lượng giác thường gặp
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GPT sau: \(4\sin\left(x+\dfrac{\pi}{3}\right)-2\sin\left(2x-\dfrac{\pi}{6}\right)=\sqrt{3}\cos x+\cos2x-2\sin x+2\)
\(sin\dfrac{x}{2}sinx-cos\dfrac{x}{2}sin^2x+1=2cos^2\left(\dfrac{pi}{4}-\dfrac{x}{2}\right)\)
giải phương trình\(\sqrt{3}cos\left(x+\dfrac{\Pi}{2}\right)+sin\left(x-\dfrac{\Pi}{2}\right)=2sin2x\)
cos²x + cos²2x + cos²3x = 1
sin²2x - sin²8x = sin\(\left(\dfrac{17\pi}{2}+10\pi\right)\)
a, cos4x + 12sin2x -1 0
b, cos4x - sin4x + cos4x 0
c, 5.(sinx + dfrac{cos3x+sin3x}{1+2sin2x} ) 3 + cos2x với mọi xinleft(0;2piright)
d, dfrac{sin3x}{3}dfrac{sin5x}{5}
e, dfrac{sin5x}{5sinx}1
f, cos23x - cos2x - cos2x 0
g, cos4x + sin4x + cos(x-dfrac{pi}{4} ) . sin(3x-dfrac{pi}{4} ) - dfrac{3}{2} 0
h, sinleft(2x+dfrac{5pi}{2}right) - 3cosleft(x-dfrac{7pi}{2}right) 1 + 2sinx với xinleft(dfrac{pi}{2};2piright)
i, 5sinx - 2 3.( 1- sinx ) . tan3x
k, ( sin2x + sqrt{3}cos2x)2 - 5 cos le...
Đọc tiếp
a, cos4x + 12sin2x -1 = 0
b, cos4x - sin4x + cos4x = 0
c, 5.(sinx + \(\dfrac{cos3x+sin3x}{1+2sin2x}\) ) = 3 + cos2x với mọi x\(\in\left(0;2\pi\right)\)
d, \(\dfrac{sin3x}{3}=\dfrac{sin5x}{5}\)
e, \(\dfrac{sin5x}{5sinx}=1\)
f, cos23x - cos2x - cos2x =0
g, cos4x + sin4x + cos(\(x-\dfrac{\pi}{4}\) ) . sin(\(3x-\dfrac{\pi}{4}\) ) - \(\dfrac{3}{2}\) = 0
h, sin\(\left(2x+\dfrac{5\pi}{2}\right)\) - 3cos\(\left(x-\dfrac{7\pi}{2}\right)\)= 1 + 2sinx với x\(\in\left(\dfrac{\pi}{2};2\pi\right)\)
i, 5sinx - 2 = 3.( 1- sinx ) . tan3x
k, ( sin2x + \(\sqrt{3}cos2x\))2 - 5 = cos \(\left(2x-\dfrac{\pi}{6}\right)\)
l, \(\dfrac{2.\left(cos^6x+sin^6x\right)-sinx.cosx}{\sqrt{2}-2sinx}=0\)
m, \(\dfrac{\left(1+sinx+cos2x\right).sin\left(x+\dfrac{\pi}{4}\right)}{1+tanx}=\dfrac{1}{\sqrt{2}}cosx\)
Mọi người giúp mình nha ! Mình cần gấp cho ngày mai 
1 + sin\(\dfrac{x}{2}\)sinx - cos\(\dfrac{x}{2}\)sin²x = 2cos²\(\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)\)
\(sin^2\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)tan^2x-cos^2\dfrac{x}{2}=0\)
GPT: \(\dfrac{4\sin^2\dfrac{x}{2}-\sqrt{3}\cos2x-1-2\cos^2\left(x-\dfrac{3\pi}{4}\right)}{\sqrt{2\cos3x+1}}=0\)
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 +\(\dfrac{cos2x}{2sinx}\)
\(\left(\dfrac{cos4x+sin2x}{cos3x+sin3x}\right)^2=2\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)+3\)