a,Cos (2x +π/5) =cos π/5
b, cos (x+π/4) = cos 2x
c, cos x = sin x
d, sin (x+ π/3)= sin 2x
e, 2sinx +1=0
k,(2cosx +1) (√2 +2sinx)
a,Cos (2x +π/5) =cos π/5
b, cos (x+π/4) = cos 2x
c, cos x = sin x
d, sin (x+ π/3)= sin 2x
e, 2sinx +1=0
k,(2cosx +1) (√2 +2sinx)
`~rai~`
\(a)cos\left(2x+\dfrac{\pi}{5}\right)=cos\dfrac{\pi}{5}\\\Leftrightarrow \left[\begin{array}{I}2x+\dfrac{\pi}{5}=\dfrac{\pi}{5}+k2\pi\\2x+\dfrac{\pi}{5}=-\dfrac{\pi}{5}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}2x=k2\pi\\2x=-\dfrac{2\pi}{5}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=k\pi\\x=-\dfrac{\pi}{5}+k\pi.\end{array}\right.\quad(k\in\mathbb{Z})\\b)cos\left(x+\dfrac{\pi}{4}\right)=cos2x\\\Leftrightarrow \left[\begin{array}{I}x+\dfrac{\pi}{4}=2x+k2\pi\\x+\dfrac{\pi}{4}=-2x+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{\pi}{4}-k2\pi\\x=-\dfrac{\pi}{12}+k\dfrac{2\pi}{3}.\end{array}\right.(k\in\mathbb{Z})\\c)cosx=sinx\\\Leftrightarrow cosx=cos\left(\dfrac{\pi}{2}-x\right)\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{\pi}{2}-2x+k2\pi\\x=-\dfrac{\pi}{2}+x+k2\pi\text{(vô nghiệm)}\end{array}\right.\\\Leftrightarrow x=\dfrac{\pi}{6}+k\dfrac{2\pi}{3}.(k\in\mathbb{Z})\\d)sin\left(x+\dfrac{\pi}{3}\right)=sin2x\\\Leftrightarrow \left[\begin{array}{I}x+\dfrac{\pi}{3}=2x+k2\pi\\x+\dfrac{\pi}{3}=\pi-2x+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{\pi}{3}-k2\pi\\x=\dfrac{2\pi}{9}+k\dfrac{2\pi}{3}.\end{array}\right.\quad(k\in\mathbb{Z})\\e)2sinx+1=0\\\Leftrightarrow sinx=-\dfrac{1}{2}\\\Leftrightarrow \left[\begin{array}{I}x=-\dfrac{\pi}{6}+k2\pi\\x=\pi-\left(-\dfrac{\pi}{6}\right)+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7\pi}{6}+k2\pi.\end{array}\right.\quad(k\in\mathbb{Z})\\k)(2cosx+1)(\sqrt{2}+2sinx)=0\\\Leftrightarrow \left[\begin{array}{I}2cosx+1=0\\\sqrt{2}+2sinx=0\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}cosx=-\dfrac{1}{2}\\sinx=-\dfrac{\sqrt{2}}{2}\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{2\pi}{3}+k2\pi\\x=-\dfrac{2\pi}{3}+k2\pi\\x=-\dfrac{\pi}{4}+k2\pi\\x=\pi-\left(-\dfrac{\pi}{4}\right)+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}=x\dfrac{2\pi}{3}+k2\pi\\x=-\dfrac{2\pi}{3}+k2\pi\\x=-\dfrac{\pi}{4}+k2\pi\\x=\dfrac{5\pi}{4}+k2\pi.\end{array}\right.(k\in\mathbb{Z})\)