Câu 17 neee $a)sin(3x+\frac\pi4)=\frac{\sqrt3}2$ $b)tan(\frac\pi5-x)=2.$ $c)cot(\frac\pi4-x)=\frac1{\sqrt3}.$ $d)cot(4x+35^0)=-1.$

Question

Accepted Answer

$\displaystyle \begin{array}{{>{\displaystyle}l}}
a,\ sin\left( 3x+\frac{\pi }{4} ight) =\frac{\sqrt{3}}{2}\
\Leftrightarrow sin\left( 3x+\frac{\pi }{4} ight) =sin\left(\frac{\pi }{3} ight)\
\Leftrightarrow \left[ \begin{array}{l l}
3x+\frac{\pi }{4} =\frac{\pi }{3} +k2\pi & \
3x+\frac{\pi }{4} =\pi -\frac{\pi }{3} +k2\pi &
\end{array} ight.\
\Leftrightarrow \left[ \begin{array}{l l}
3x=\frac{\pi }{12} +k2\pi & \
3x=\frac{5\pi }{12} +k2\pi &
\end{array} ight.\
\Leftrightarrow \left[ \begin{array}{l l}
x=\frac{\pi }{36} +\frac{2k}{3} \pi & \
x=\frac{5\pi }{36} +\frac{2k}{3} \pi &
\end{array} ight.
\end{array}$
$\displaystyle \begin{array}{{>{\displaystyle}l}}
b,\ tan\left(\frac{\pi }{5} -x ight) =2\
\Leftrightarrow \frac{\pi }{5} -x=arctan( 2) +k\pi \
\Leftrightarrow x=\frac{\pi }{5} -arctan( 2) +k\pi
\end{array}$
$\displaystyle \begin{array}{{>{\displaystyle}l}}
c,\ cot\left(\frac{\pi }{4} -x ight) =\frac{1}{\sqrt{3}}\
\Leftrightarrow cot\left(\frac{\pi }{4} -x ight) =cot\left(\frac{\pi }{3} ight)\
\Leftrightarrow \frac{\pi }{4} -x=\frac{\pi }{3} +k\pi \
\Leftrightarrow x=\frac{\pi }{4} -\frac{\pi }{3} -k\pi \
\Leftrightarrow x=\frac{-\pi }{12} -k\pi
\end{array}$
$\displaystyle \begin{array}{{>{\displaystyle}l}}
d,\ cot\left( 4x+35^{0} ight) =-1\
\Leftrightarrow cot\left( 4x+35^{0} ight) =cot\left( 135^{0} ight)\
\Leftrightarrow 4x+35^{0} =135^{0} +180^{0} .k\
\Leftrightarrow 4x=100^{0} +180^{0} k\
\Leftrightarrow x=25^{0} +45^{0} .k
\end{array}$