Giải phương trình
a. √3tan (x+π/4)+1=0.
b. Sin^2 2x+5sin2x-4=0.
C. Sin x-√3cos=√2.
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1 câu trả lời 3169
\(\begin{array}{l}
a)\sqrt 3 tan\left( {x + \frac{\pi }{4}} \right) + 1 = 0\\
\Leftrightarrow \tan (x + \frac{\pi }{4}) = \frac{{ - 1}}{{\sqrt 3 }}\\
\Leftrightarrow \tan (x + \frac{\pi }{4}) = \tan \frac{{ - \pi }}{6}\\
\Leftrightarrow x + \frac{\pi }{4} = \frac{{ - \pi }}{6} + k\pi \\
\Leftrightarrow x = \frac{{ - 5\pi }}{{12}} + k\pi (k \in Z)\\
b)Si{n^2}2x + 5sin2x - 4 = 0\\
\Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{\sin 2x = \frac{{ - 5 + \sqrt {41} }}{2}}\\
{\sin 2x = \frac{{ - 5 - \sqrt {41} }}{2}(ktm)}
\end{array}} \right.\\
\Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{x = \frac{{\arcsin \frac{{ - 5 + \sqrt {41} }}{2}}}{2} + k\pi }\\
{x = \frac{{\pi - \arcsin \frac{{ - 5 + \sqrt {41} }}{2}}}{2} + k\pi }
\end{array}} \right.(k \in Z)\\
c)\;Sin{\rm{ }}x - \sqrt 3 cos = \sqrt 2 \\
\Leftrightarrow \frac{1}{2}\sin x - \frac{{\sqrt 3 }}{2}\cos x = \frac{{\sqrt 2 }}{2}\\
\Leftrightarrow \sin (x - \frac{\pi }{3}) = \sin \frac{\pi }{4}\\
\Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{x - \frac{\pi }{3} = \frac{\pi }{4} + k2\pi }\\
{x - \frac{\pi }{3} = \pi - \frac{\pi }{4} + k2\pi }
\end{array}} \right.\\
\Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{x = \frac{{7\pi }}{{12}} + k2\pi }\\
{x = \frac{{13\pi }}{{12}} + k2\pi }
\end{array}} \right.(k \in Z)
\end{array}\)
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