\(1-2cos^2x-sinx=0\)

\(\Leftrightarrow1-2\left(1-sin^2x\right)-sinx=0\)

\(\Leftrightarrow2sin^2x-sinx-1=0\Rightarrow\left[{}\begin{matrix}sinx=1\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Rightarrow x=\left\{\dfrac{\pi}{2};\dfrac{7\pi}{6};\dfrac{11\pi}{6};\dfrac{5\pi}{2}\right\}\)

\(\Rightarrow\sum x=6\pi\)

\(\dfrac{3\pi}{2}< a< 2\pi\Rightarrow sina< 0\)

\(\Rightarrow sina=-\sqrt{1-cos^2a}=-\sqrt{1-\left(\dfrac{3}{5}\right)^2}=-\dfrac{4}{5}\)

\(\Rightarrow sin2a=2sina.cosa=2.\left(-\dfrac{4}{5}\right).\left(\dfrac{3}{5}\right)=-\dfrac{24}{25}\)

Câu sau có nhầm đề ko nhỉ?

\(sin\left(\pi-\dfrac{\pi}{3}\right)=sin\left(\dfrac{2\pi}{3}\right)=\dfrac{\sqrt{3}}{2}\)

\(P=sin^22x-\left[2sin\dfrac{x}{2}cos\dfrac{x}{2}\left(cos^4\dfrac{x}{2}-sin^4\dfrac{x}{2}\right)\right]^2\)

\(=sin^22x-\left[sinx\left(cos^2\dfrac{x}{2}-sin^2\dfrac{x}{2}\right)\left(cos^2\dfrac{x}{2}+sin^2\dfrac{x}{2}\right)\right]^2\)

\(=sin^22x-\left[sinx.cosx.1\right]^2\)

\(=sin^22x-\left[\dfrac{1}{2}sin2x\right]^2\)

\(=\dfrac{3}{4}sin^22x=\dfrac{3}{4}\left(1-cos^22x\right)=\dfrac{3}{4}\left(1-\dfrac{1}{4}\right)=\dfrac{9}{16}\)

cảm ơn bạn nhìu :3

D=sin(pi+x)+sinx+cot(pi-x)+tan(pi/2-x)

=-sinx+sinx-cotx+cotx=0

Ta có : sin x =3/5 suy ra 5sin x = 3 

25sin²x=9 

25(1-cos²)=9

25cos²=16

5cos x =4

cos x = 4/5 . (1)

Thay (1) và sin x =3/5 vào M , ta được :

M=29/5

Sửa đề: \(180^0 và \(\sin x=-\frac45\)

\(180^0

=>sin x<0; cos x<0

Ta có: \(\sin^2x+cos^2x=1\)

=>\(cos^2x=1-\left(-\frac45\right)^2=1-\frac{16}{25}=\frac{9}{25}\)

mà cos x <0

nên \(cosx=-\frac35\)

\(cos\left(x-45^0\right)=cosx\cdot cos45^0+\sin x\cdot\sin45^0\)

\(=-\frac35\cdot\frac{1}{\sqrt2}+\frac{-4}{5}\cdot\frac{1}{\sqrt2}=\frac{-7}{5\sqrt2}=\frac{-7\sqrt2}{10}\)

Sửa đề: cosa=3/5

3pi/2<a<2pi

=>sin a<0

\(sin^2a+cos^2a=1\)

=>\(sin^2a=1-\dfrac{9}{25}=\dfrac{16}{25}\)

mà sin a<0

nên sina =-4/5

tan a=-4/5:3/5=-4/3

cot a=1:(-4/3)=-3/4