Question

Answer 1

a.

\(sin2x=1\)

\(\Leftrightarrow2x=\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\) (\(k\in Z\) ), mấy câu sau sẽ ko ghi lại cái này, bạn tự ghi

b.

\(\sqrt{3}tan\left(x-\dfrac{\pi}{4}\right)=1\)

\(\Leftrightarrow tan\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{3}}\)

\(\Leftrightarrow x-\dfrac{\pi}{4}=\dfrac{\pi}{6}+k\pi\)

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

Answer 2

c.

\(\sqrt{3}cosx-sinx=\sqrt{2}\)

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}cosx-\dfrac{1}{2}sinx=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow cosx.cos\left(\dfrac{\pi}{6}\right)-sinx.sin\left(\dfrac{\pi}{6}\right)=cos\left(\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow cos\left(x+\dfrac{\pi}{6}\right)=cos\left(\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{6}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{6}=-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+k2\pi\\x=-\dfrac{5\pi}{12}+k2\pi\end{matrix}\right.\)

Answer 3

d.

\(cos2x-3cosx=4cos^2\dfrac{x}{2}\)

\(\Leftrightarrow2cos^2x-1-3cosx=4\left(\dfrac{1+cosx}{2}\right)\)

\(\Leftrightarrow2cos^2x-5cosx-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=3>1\left(loại\right)\\cosx=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow x=\pm\dfrac{2\pi}{3}+k2\pi\)

Answer 4

e.

\(4+3sinx+sin^3x=3cos^2x+cos^6x\)

\(\Leftrightarrow4+3sinx+sin^3x=3-3sin^2x+cos^6x\)

\(\Leftrightarrow sin^3x+3sin^2x+3sinx+1=cos^6x\)

\(\Leftrightarrow\left(sinx+1\right)^3=\left(cos^2x\right)^3\)

\(\Leftrightarrow sinx+1=cos^2x\)

\(\Leftrightarrow sinx+1-cos^2x=0\)

\(\Leftrightarrow sinx+sin^2x=0\)

\(\Rightarrow\left[{}\begin{matrix}sinx=0\\sinx=-1\end{matrix}\right.\)

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

Answer 5

f.

\(cos2x+5=2\left(2-cosx\right)\left(sinx-cosx\right)\)

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

\(\Leftrightarrow2\left(sinx-cosx\right)-sinx.cosx-2=0\)

Đặt \(sinx-cosx=t\in\left[-\sqrt{2};\sqrt{2}\right]\)

\(\Rightarrow sinx.cosx=\dfrac{1-t^2}{2}\)

\(\Rightarrow2t-\dfrac{1-t^2}{2}-2=0\)

\(\Leftrightarrow t^2+4t-5=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-5\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow sinx-cosx=1\)

\(\Leftrightarrow\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=1\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)

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

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