Question

Answer 1

a, \(2cos2x+\sqrt{3}=0\)

\(\Leftrightarrow cos2x=-\dfrac{\sqrt{3}}{2}\)

\(\Leftrightarrow2x=\pm\dfrac{5\pi}{6}+k2\pi\)

\(\Leftrightarrow x=\pm\dfrac{5\pi}{12}+k\pi\)

Answer 2

b, \(2cos^2x-3cosx+1=0\)

\(\Leftrightarrow\left(cosx-1\right)\left(2cosx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pm\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

Answer 3

c, \(2sin^2x+cosx-1=0\)

\(\Leftrightarrow2cos^2x-cosx-1=0\)

\(\Leftrightarrow\left(cosx-1\right)\left(2cosx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pm\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)

Answer 4

d, \(cos4x-\sqrt{3}sin4x=2sin3x\)

\(\Leftrightarrow\dfrac{1}{2}cos4x-\dfrac{\sqrt{3}}{2}sin4x=sin3x\)

\(\Leftrightarrow cos\left(4x-\dfrac{\pi}{3}\right)=cos\left(\dfrac{\pi}{2}-3x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-\dfrac{\pi}{3}=\dfrac{\pi}{2}-3x+k2\pi\\4x-\dfrac{\pi}{3}=-\dfrac{\pi}{2}+3x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}7x=\dfrac{5\pi}{6}+k2\pi\\x=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)

Answer 5

e, \(3cos^2x-4sinx.cosx+sin^2x=1\)

\(\Leftrightarrow3cos^2x-4sinx.cosx+sin^2x-sin^2x-cos^2x=0\)

\(\Leftrightarrow2cos^2x-4sinx.cosx=0\)

\(\Leftrightarrow2cosx\left(cosx-2sinx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx-2sinx=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\cos\left(x+arccos\dfrac{1}{\sqrt{5}}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x+arccos\dfrac{1}{\sqrt{5}}=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{\pi}{2}-arccos\dfrac{1}{\sqrt{5}}+k\pi\end{matrix}\right.\)

Answer 6

f, \(sin2x-sin\left(x+\dfrac{2\pi}{5}\right)=0\)

\(\Leftrightarrow2cos\left(\dfrac{3x}{2}+\dfrac{\pi}{5}\right).sin\left(\dfrac{x}{2}-\dfrac{\pi}{5}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(\dfrac{3x}{2}+\dfrac{\pi}{5}\right)=0\\sin\left(\dfrac{x}{2}-\dfrac{\pi}{5}\right)=0\end{matrix}\right.\)

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