a: \(2\cdot cos\left(2x+\frac{\pi}{6}\right)=\sqrt2\)
=>\(cos\left(2x+\frac{\pi}{6}\right)=\frac{\sqrt2}{2}\)
=>\(\left[\begin{array}{l}2x+\frac{\pi}{6}=\frac{\pi}{4}+k2\pi\\ 2x+\frac{\pi}{6}=-\frac{\pi}{4}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=\frac{\pi}{4}-\frac{\pi}{6}+k2\pi=\frac{\pi}{12}+k2\pi\\ 2x=-\frac{\pi}{4}-\frac{\pi}{6}+k2\pi=-\frac{5}{12}\pi+k2\pi\end{array}\right.\)
=>\(\left[\begin{array}{l}x=\frac{\pi}{24}+k\pi\\ x=-\frac{5}{24}\pi+k\pi\end{array}\right.\)
TH1: \(x=\frac{\pi}{24}+k\pi\)
\(x\in\left(-\frac56\pi;\frac32\pi\right)\)
=>\(\frac{\pi}{24}+k\pi\in\left(-\frac56\pi;\frac32\pi\right)\)
=>\(k+\frac{1}{24}\in\left(-\frac56;\frac32\right)\)
=>\(k\in\left(-\frac56-\frac{1}{24};\frac32-\frac{1}{24}\right)\)
=>k∈(-21/24;35/24)
=>k∈{0;1}
Khi k=0 thì \(x=\frac{\pi}{24}+0\cdot\pi=\frac{\pi}{24}\)
Khi k=1 thì \(x=\frac{\pi}{24}+\pi=\frac{25}{24}\pi\)
TH2: \(x=-\frac{5}{24}\pi+k\pi\)
\(x\in\left(-\frac56\pi;\frac32\pi\right)\)
=>\(\frac{-5\pi}{24}+k\pi\in\left(-\frac56\pi;\frac32\pi\right)\)
=>\(k-\frac{5}{24}\in\left(-\frac56;\frac32\right)\)
=>\(k\in\left(-\frac56+\frac{5}{24};\frac32+\frac{5}{24}\right)\)
=>k∈(-19/24;41/24)
=>k∈{0;1}
Khi k=0 thì \(x=-\frac{5}{24}\pi+0\cdot\pi=-\frac{5}{24}\pi\)
Khi k=1 thì \(x=-\frac{5}{24}\pi+\pi=\frac{19}{24}\pi\)
b: \(\tan\left(2x-15^0\right)=1\)
=>\(2x-15^0=45^0+k\cdot180^0\)
=>\(2x=60^0+k\cdot180^0\)
=>\(x=30^0+k\cdot90^0\)
mà -180 độ<x<90 độ
nên x∈{-150 độ; -60 độ; 30 độ}
