tan(x+pi/2)=1
\(\Leftrightarrow x+\dfrac{\Pi}{2}=\dfrac{\Pi}{4}+k\Pi\)
\(\Leftrightarrow x=-\dfrac{\Pi}{4}+k\Pi\)
\(sin\left(2x-\dfrac{\Pi}{6}\right)=sin\left(-\dfrac{\Pi}{2}+2k\Pi-\dfrac{\Pi}{6}\right)\)
\(=sin\left(k2\Pi-\dfrac{2}{3}\Pi\right)=sin\left(-\dfrac{2}{3}\Pi\right)=-\dfrac{\sqrt{3}}{2}\)
Cho \(tan\left(x+\frac{\Pi}{2}\right)-1=0\) . Tính \(sin\left(2x-\frac{\Pi}{6}\right)\) .
A . \(sin\left(2x-\frac{\Pi}{6}\right)=-\frac{1}{2}\)
B . \(sin\left(2x-\frac{\Pi}{6}\right)=\frac{\sqrt{3}}{2}\)
C . \(sin\left(2x-\frac{\Pi}{6}\right)=-\frac{\sqrt{3}}{2}\)
D . \(sin\left(2x-\frac{\Pi}{6}\right)=\frac{1}{2}\)
Trình bày bài giải chi tiết rồi ms chọn đáp án nha các bạn .
1)giải pt a)√2 cos2x-1=0
b) sinx =cos3x
c) cos (x+π/3) +sin(3x+π/4)=0
d)tan 2x = cot (x+π/4)
e) sin x = √3 cos x
f) tan^2(π/3-2x)-3=0
cho \(tan\left(x+\frac{\pi}{2}\right)-1=0.\)Tính \(sin\left(2x-\frac{\pi}{6}\right)\)
bài 1: a) \(sin\left(2x+\dfrac{\pi}{6}\right)+sin\left(x-\dfrac{\pi}{3}\right)=0\)
b) \(sin\left(2x-\dfrac{\pi}{3}\right)-cos\left(x+\dfrac{\pi}{3}\right)=0\)
c) \(sin\left(2x+\dfrac{\pi}{3}\right)+cos\left(x-\dfrac{\pi}{6}\right)=0\)
Xét tính chẵn, lẻ của các hàm số
1,\(y=cosx+sin^2x\)
2,\(y=sinx+cosx\)
3,\(y=tanx+2sinx\)
4,\(y=tan2x-sin3x\)
5,\(sin2x+cosx\)
6,\(y=cosx.sin^2x-tan^2x\)
7,\(y=cos\left(x-\dfrac{\pi}{4}\right)+cos\left(x+\dfrac{\pi}{4}\right)\)
8,\(y=\dfrac{2+cosx}{1+sin^2x}\)
9,\(y=\left|2+sinx\right|+\left|2-sinx\right|\)
Giải các phương trình:
a) \(\left|\sin x+\dfrac{1}{2}\right|=\dfrac{1}{2}\)
b) \(\tan^2\left(x+\dfrac{\pi}{6}\right)=3\)
c) \(2\sin\left(4x-\dfrac{\pi}{3}\right)-1=0\)
Giải phương trình:
1) \(cos\left(2x + \dfrac{\pi}{6}\right) = cos\left(\dfrac{\pi}{3} - 3x\right)\)
2) \(sin\left(2x + \dfrac{\pi}{6}\right) = sin\left(\dfrac{\pi}{3} - 3x\right)\)
tìm tập xác định của hàm số lượng giác sau
a)\(y=\dfrac{tan\left(2x-\dfrac{\pi}{4}\right)}{\sqrt{1-sin\left(x-\dfrac{\pi}{8}\right)}}\)
b)\(y=\dfrac{tan\left(x-\dfrac{\pi}{4}\right)}{1-cos\left(x+\dfrac{\pi}{3}\right)}\)
c)\(y=\dfrac{3}{cosx-cos3x}\)
d)\(y=\dfrac{4}{sin^2x-cos^2x}\)
e)\(y=\dfrac{1+cot\left(\dfrac{\pi}{3}+x\right)}{tan^2\left(3x-\dfrac{\pi}{4}\right)}\)
1. sin10x=1
2. tan(\(\frac{x}{3}\)+\(\frac{\pi}{3}\))=\(\frac{\sqrt{3}}{3}\)
3. sin(2x+\(\frac{\pi}{3}\))+sinx=0
4.8cos2x.sin2x.cos4x=\(\sqrt{2}\)
5.cot2x=cot(x+\(\frac{\pi}{2}\))
6. tan(x+\(\frac{\pi}{3}\))+cot2x=0
7. cosx.sin(\(\frac{\pi}{2}\)+6x)+cos(\(\frac{\pi}{2}\)-x).sin6x=1
8. cos(2x+\(\frac{\pi}{3}\))+cos(x-\(\frac{\pi}{3}\))=0
